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Problems 
Solutions 








Shepherd 







52- 




X- 



ARNOLD 



Surveying 

Problems and 

Solutions 



F. A. Shepherd 



■£ 



■»ii^,^iiA 



* 



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Thfs new book gives a presentation 
concentrating on mathematical problems, 
an aspect of the subject which usually 
causes most difficulty. 

Summaries of basic theory are followed 
by worked examples and selected exer- 
cises. The book covers three main 
branches of surveying: measurement, 
surveying techniques and industrial appli- 
cations. It is a book concerned mainly 
with engineering surveying as applied, 
for example, in the construction and 
mining industries. 



Contents 

Linear Measurement 

Surveying Trigonometry 

Co-ordinates 

Instrumental Optics 

Levelling 

Traverse Surveys 

Tacheometry 

Dip and Fault Problems 

Areas 

Volumes 

Circular Curves 

Vertical and Transition Curves 

Values in both imperial and metric (S. 
units are given in the problems 



Edward Arnold 



80s. net 



Edward Arnold (Publishers) Ltd., 
41 Maddox Street, London, W.I. 

Printed in Great Britain 






SURVEYING 

PROBLEMS & 

SOLUTIONS 




Shop l>ord 


1 



Surveying Problems 
and Solutions 

F. A. Shepherd c.En g ., A.R.i.c.s.,M.i.Min.E. 

Senior Lecturer in Surveying 

Nottingham Regional College of Technology 



London. Edward Arnold (Publishers) Ltd. 



HARRIS Co<...jE | 
PISTON ! 



I 

© F.A. Shepherd 1968 






First published 1968 

Boards edition SBN: Q7 131 3198 5t 
Paper edition SBN: 7131 3199 3 



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Printed in Great Britain by Bookprint Ltd., Crawley, Sussex 



PREFACE 

This book is an attempt to deal with the basic mathematical aspects 
of 'Engineering Surveying', i.e. surveying applied to construction and 
mining engineering projects, and to give guidance on practical methods 
of solving the typical problems posed in practice and, in theory, by the 
various examining bodies. 

The general approach adopted is to give a theoretical analysis of 
each topic, followed by worked examples and, finally, selected exer- 
cises for private study. Little claim is made to new ideas, as the 
ground covered is elementary and generally well accepted. It is hoped 
that the mathematics of surveying, which so often causes trouble to 
beginners, is presented in as clear and readily understood a manner as 
possible. The main part of the work of the engineering surveyor, civil 
and mining engineer, and all workers in the construction industry is 
confined to plane surveying, and this book is similarly restricted. 

It is hoped that the order of the chapters provides a natural 
sequence, viz.: 

(a) Fundamental measurement 

(i) Linear measurement in the horizontal plane, 
(ii) Angular measurement and its relationship to linear values, 

i.e. trigonometry, 
(iii) Co-ordinates as a graphical and mathematical tool. 

(b) Fundamental surveying techniques 
(i) Instrumentation. 

(ii) Linear measurement in the vertical plane, i.e. levelling, 
(iii) Traversing as a control system, 
(iv) Tacheometry as a detail and control system. 

(c) Industrial applications 

(i) Three-dimensional aspects involving inclined planes, 
(ii) Mensuration, 
(iii) Curve surveying. 

Basic trigonometry is included, to provide a fundamental mathe- 
matical tool for the surveyor. It is generally found that there is a 
deficiency in the student's ability to apply numerical values to trigo- 
nometrical problems, particularly in the solution of triangles, and it is 
hoped that the chapter in question shows that more is required than the 
sine and cosine formulae. Many aspects of surveying, e.g. errors in 
surveying, curve ranging, etc. require the use of small angles, and the 
application of radians is suggested. Few numerical problems are posed 
relating to instrumentation, but it is felt that a knowledge of basic 



physical properties affords a more complete understanding of the con- 
struction and use of instruments. To facilitate a real grasp of the sub- 
ject, the effects of errors are analysed in all sections. This may 
appear too advanced for students who are not familiar with the element- 
ary calculus, but it is hoped that the conclusions derived will be 
beneficial to all. 

With the introduction of the Metric System in the British Isles and 
elsewhere, its effect on all aspects of surveying is pin-pointed and 
conversion factors are given. Some examples are duplicated in the 
proposed units based on the International System (S.I.) and in order to 
give a 'feel' for the new system, during the difficult transition period, 
equivalent S.I. values are given in brackets for a few selected examples. 

The book is suitable for all students in Universities and Technical 
Colleges, as well as for supplementary postal tuition, in such courses 
as Higher National Certificates, Diplomas and Degrees in Surveying, 
Construction, Architecture, Planning, Estate Management, Civil and 
Mining Engineering, as well as for professional qualification for the 
Royal Institution of Chartered Surveyors, the Institution of Civil 
Engineers, the Incorporated Association of Architects and Surveyors, 
the Institute of Quantity Surveyors, and the Institute of Building. 

ACKNOWLEDGMENTS 

I am greatly indebted to the Mining Qualifications Board (Ministry 
of Power) and the Controller of H.M. Stationery Office, who have given 
permission for the reproduction of examination questions. My thanks 
are also due to the Royal Institution of Chartered Surveyors, the 
Institution of Civil Engineers, to the Senates of the Universities of 
London and Nottingham, to the East Midlands Educational Union and 
the Nottingham Regional College of Technology, all of whom have 
allowed their examination questions to be used. 

My special thanks are due to many of my colleagues at Nottingham, 
but especially to Messrs. J. H. Ball, A.R.I.C.S., A.I.A.S., 
A.M.I.Min.E., A. Eaton, B.Sc., C.Eng., A.M.I.C.E., A.M.B.I.M., 
G. M. Lewis, B.Sc, Ph.D., M. B. Pate, M.Sc, A. A. Payne, B.Sc, 
C. Rayner, B.Sc, A.R.I.C.S., R. Robb, A.R.I.C.S., A.M.I.Min.E., 
D.B. Shaw, B.Sc, and J. P. Withers, B.Sc, C.Eng., A.M.I.C.E., all of 
whom have offered advice and help in checking the text 

The ultimate responsibility for the accuracy is, of course, my own. 
I am very conscious that, even with the most careful checking, it is not 
to be expected that every mistake has been eliminated, and I can only 
ask readers if they will kindly bring any errors to my notice. 
Nottingham F. A. SHEPHERD 

1968 





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VI 



CONVERSION FACTORS (B) 



(Ref 




Changing to the Metric System, H.M.S.O., 1967) 


Length 










1 mile 


= 


1-609 34 km 


1km = 


0-621371 mile 


1 furlong 


= 


0-201 168 km 






1 chain 


= 


20-116 8 m 






1yd 


= 


0-914 4 m 


lm 


1-093 61 yd 


1ft 


= 


0-304 8 m 






lin. 


= 


2-54 cm 


1 cm = 


0-393701 in. 


1 fathom 


= 


1-828 8 m 






llink 


= 


0-201 168 m 






Area 










1 sq. mile 


= 


2-58999 km 2 






1 acre 


= 


4046-86 m 2 


1 km 2 = 


247-105 acres 


1 rood 


= 


1011-71 m 2 






1yd 2 


= 


0-836 127 m 2 


lm 2 = 


1-19599 yd 2 


1ft 2 


= 


0-092903 m 2 






lin 2 


= 


6-4516 cm 2 


1 cm 2 = 


0-15500 in 2 


1 sq. chain 


= 


404-686 m 2 






Volume 










1yd 3 


= 


0-764555 m 3 


lm 3 = 


1-30795 yd 3 


1 ft 3 


= 


0-028 316 8 m 3 


lm 3 = 


35-314 7 ft 3 


lin 3 


= 


16-387 1 cm 3 


1 cm 3 = 


0-061023 7 in 3 


lgal 


= 


0-004546 09 m 3 








= 


4-54609 litre 


1 litre = 


0-2200 gal 


Velocity 










1 mile/h 


= 


1-609 34 km/h 


1 km/h = 


0-621371 m. p. r 


lft/s 


= 


0-3048 m/s 


1 m/s = 


3-28084 ft/s 


Acceleration 










1 ft/s 2 


= 


0-3048 m/s 2 


1 m/s 2 = 


3-280 84 ft/s 2 


Mass 










1 ton 


= 


1016-05 kg 






1 cwt 


= 


50-8023 kg 






lib 


= 


0-453 59237 kg 


1kg = 


2-20462 lb 



Vll 

Mass per unit length 
1 lb/ft = 1-488 16 kg/m 

Mass per unit area 
lib/ft 2 = 4-88243 kg/m 2 

Density 

1 ton/yd 3 = 1328-94 kg/m 3 

1 lb/ft 3 = 16-018 5 kg/m 3 1 kg/m 3 = 0-062428 lb/ft 3 

1 lb/gal = 99-776 3 kg/m 3 

0-09978 kg/1 
Force 

Hbf = 4-448 22 N IN = 0-224 809 lbf 

Ikgf = 9-80665 N 1 kgf = 2-20462 lbf 

Force (weight) /unit length 
1 lbf/ft = 14-593 9 N'm 

Pressure 

1 lbf/ft 2 = 47-880 3 N/m 2 1 N/m 2 = 0-000 145 038 lbf/in 2 

1 lbf /in 2 = 6894-76 N/m 2 

1 kgf /cm 2 = 98-066 5 kN/m 2 

lkgf/m 2 = 9-80665 N/m 2 

Standard gravity 

32-1740 ft/s 2 = 9-80665 m/s 2 

N.B. lib = 0-453 592 kg 

1 lbf = 0-453 592 x 9-80665 = 4-448 22 N 

1 newton (N) unit of force = that force which applied to a mass 
of 1 kg gives an acceleration of 1 m/s 2 . 



Vlll 

CONTENTS 

Chapter Page 

1 LINEAR MEASUREMENT 1 

1.1 The basic principles of surveying 1 

1.2 General theory of measurement 2 

1.3 Significant figures in measurement and computation 3 

1.4 Chain surveying 6 

1.41 Corrections to ground measurements 6 

1.42 The maximum length of offsets from chain lines 13 

1.43 Setting out a right angle by chain 15 

1.44 To find the point on the chain line which produces 

a perpendicular from a point outside the line 16 

1.45 Obstacles in chain surveying 17 
Exercises 1(a) 22 

1.5 Corrections to be applied to measured lengths 23 

1.51 Standardisation 23 

1.52 Correction for slope 23 

1.53 Correction for temperature 26 

1.54 Correction for tension 27 

1.55 Correction for sag 32 

1.56 Reduction to mean sea level 38 

1.57 Reduction of ground length to grid length 39 

1.6 The effect of errors in linear measurement 45 

1.61 Standardisation 45 

1.62 Malalignment and deformation of the tape 45 

1.63 Reading or marking the tape 46 

1.64 Errors due to wrongly recorded temperature 46 

1.65 Errors due to variation from the recorded value of 
tension 47 

1.66 Errors from sag 48 

1.67 Inaccurate reduction to the horizontal 51 

1.68 Errors in reduction from height above or below 

mean sea level 52 

1.69 Errors due to the difference between ground and 

grid distances 52 

Exercises 1(b) 52 

2 SURVEYING TRIGONOMETRY 57 

2.1 Angular measurement 57 

2.11 The degree system 57 



2.12 Trigonometrical ratios 58 

2.13 Complementary angles 60 

2.14 Supplementary angles 60 

2.15 Basis of tables of trigonometrical functions 63 

2.16 Trigonometric ratios of common angles 64 

2.17 Points of the compass 65 

2. 18 Easy problems based on the solution of the right- 
angled triangle 67 
Exercises 2(a) 71 

2.2 Circular measure 72 

2.21 The radian 72 

2.22 Small angles and approximations 73 

2.3 Trigonometrical ratios of the sums and differences of 

two angles 77 

2.4 Transformation of products and sums 79 

2.5 The solution of triangles 80 

2.51 Sine rule 80 

2.52 Cosine rule 81 

2.53 Area of a triangle 82 

2.54 Half-angle formulae 82 

2.55 Napier's tangent rule 83 

2.56 Problems involving the solution of triangles 83 

2.6 Heights and distances 91 

2.61 To find the height of an object having a vertical face 91 

2.62 To find the height of an object when its base 

is inaccessible 92 

2.63 To find the height of an object above the ground 
when its base and top are visible but not 

accessible 95 

2.64 To find the length of an inclined object on the top 

of a building 98 

2.65 To find the height of an object from three angles 

of elevation only 100 

2.66 The broken base line problem 102 
2; 67 To find the relationship between angles in the 

horizontal and inclined planes 106 

Exercises 2(b) 108 

CO-ORDINATES 112 

3.1 Polar co-ordinates 112 

3.11 Plotting to scale 113 

3.12 Conversion of the scales 113 

3.13 Scales in common use 114 



3.14 Plotting accuracy 114 

3.15 Incorrect scale problems 114 

3.2 Bearings 115 

3.21 True north 115 

3.22 Magnetic north 115 

3.23 Grid north 116 

3.24 Arbitrary north 116 

3.25 Types of bearing 117 

3.26 Conversion of horizontal angles into bearings 121 

3.27 Deflection angles 124 
Exercises 3(a) 126 

3.3 Rectangular co-ordinates 127 

3.31 Partial co-ordinates, AE, AN 128 

3.32 Total co-ordinates 128 
Exercises 3(b) (Plotting) 131 

3.4 Computation processes 133 

3.41 Computation by logarithms 134 

3.42 Computation by machine 134 

3.43 Tabulation process 135 

3.44 To obtain the bearing and distance between two 

points given their co-ordinates 136 

3.5 To find the co-ordinates of the intersection of two lines 146 

3.51 Given their bearings from two known co-ordinate 

stations 146 

3.52 Given the length and bearing of a line AB and all 

the angles A, B and C 149 

Exercises 3(c) (Boundaries) 157 

3.6 Transposition of grid 158 

3.7 The National Grid Reference system 160 

Exercises 3(d) (Co-ordinates) 163 

Appendix (Comparison of Scales) 169 

4 INSTRUMENTAL OPTICS 170 

4.1 Reflection at plane surfaces 170 

4.11 Laws of reflection 170 

4.12 Deviation by successive reflections on two inclined 
mirrors 170 

4.13 The optical square 171 

4.14 Deviation by rotating the mirror 171 

4.15 Principles of the sextant 172 

4.16 Use of the true horizon 174 

4.17 Artificial horizon 175 

4.18 Images in plane mirrors 176 

4.19 Virtual and real images 177 



XI 



4.2 Refraction at plane surfaces 177 

4.21 Laws of refraction 177 

4.22 Total internal reflection 177 

4.23 Relationships between refractive indices 178 

4.24 Refraction through triangular prisms 179 

4.25 Instruments using refraction through prisms 180 
Exercises 4(a) 184 

4.3 Spherical mirrors 184 

4.31 Concave or converging mirrors 184 

4.32 Convex or diverging mirrors 186 

4.33 The relationship between object and image in 

curved mirrors 186 

4.34 Sign convention lg7 

4.35 Derivation of formulae Igg 

4.36 Magnification in spherical mirrors 190 

4.4 Refraction through thin lenses 191 

4.41 Definitions 191 

4.42 Formation of images 192 

4.43 The relationship between object and image in 

a thin lens 193 

4.44 Derivation of formulae 193 

4.45 Magnification in thin lenses 195 

4.5 Telescopes 196 

4.51 Kepler's astronomical telescope 196 

4.52 Galileo's telescope 196 

4.53 Eyepieces I97 

4.54 The internal focussing telescope 198 

4.55 The tacheometric telescope (external focussing) 201 

4.56 The anallatic lens 203 

4.57 The tacheometric telescope (internal focussing) 207 

4.6 Instrumental errors in the theodolite 210 

4.61 Eccentricity of the horizontal circle 210 

4.62 The line of collimation not perpendicular to the 

trunnion axis 213 

4.63 The trunnion axis not perpendicular to the 

vertical axis 215 

4.64 Vertical axis not truly vertical 217 
4-65 Vertical circle index error 219 

4.7 The auxiliary telescope 228 

4.71 Side telescope 228 

4.72 Top telescope 233 



Xll 



4.8 Angular error due to defective centring of 

the theodolite 234 

4.9 The vernier 237 

4.91 Direct reading vernier 237 

4.92 Retrograde vernier 238 

4.93 Special forms used in vernier theodolites 238 
4-94 Geometrical construction of the vernier 

scale 238 

Exercises 4(b) 240 

LEVELLING 244 

5. 1 Definitions 244 

5.2 Principles 245 

5.3 Booking, of readings 246 

5.31 Method 1, rise and fall 246 

5.32 Method 2, height of collimation 247 
Exercises 5 (a) (Booking) 254 

5.4 Field testing of the level 257 

5.41 Reciprocal levelling method 257 

5.42 Two-peg method 259 
Exercises 5 (b) (Adjustment) 264 

5.5 Sensitivity of the bubble tube 267 

5.51 Field test 267 

5.52 O-E correction 268 

5.53 Bubble scale correction 268 
Exercises 5(c) (Sensitivity) 270 

5.54 Gradient screws (tilting mechanism) 271 

5.6 The effect of the earth's curvature and atmospheric 
refraction 272 

5.61 The earth's curvature 272 

5-62 Atmospheric refraction 273 

5.63 The combined effect of curvature and refraction 273 
Exercises 5(d) (Curvature and refraction) 275 

5.64 Intervisibility 275 
Exercises 5 (e) (Intervisibility) 277 

5.65 Trigonometrical levelling 278 

5.7 Reciprocal levelling 279 
5.71 The use of two instruments 281 

Exercises 5(f) (Reciprocal levelling) 282 

5.8 Levelling for construction 283 
5.81 Grading of constructions 283 



Xlll 



5.82 The use of sight rails and boning 

(or travelling) rods 284 

5.83 The setting of slope stakes 286 
Exercises 5(g) (Construction levelling) 288 
Exercises 5 (h) (General) 289 

TRAVERSE SURVEYS 298 

6. 1 Types of traverse 298 

6.11 Open 298 

6.12 Closed 298 

6.2 Methods of traversing 299 

6.21 Compass traversing 300 

6.22 Continuous azimuth method 301 

6.23 Direction method 302 

6. 24 Separate angular measurement 304 
Exercises 6(a) 304 

6.3 Office tests for locating mistakes in traversing 306 

6.31 A mistake in the linear value of one line 306 

6.32 A mistake in the angular value at one station 307 

6.33 When the traverse is closed on to fixed points 

and a mistake in the bearing is known to exist 307 

6.4 Omitted measurements in closed traverses 308 

6.41 Where the bearing of one line is missing 308 

6.42 Where the length of one line is missing 309 
6-43 Where the length and bearing of a line 

are missing 309 

6.44 Where the bearings of two lines are missing 309 

6.45 Where two lengths are missing 314 

6.46 Where the length of one line and the bearing 

of another line are missing 315 

Exercises 6(b) (Omitted values) 316 

6.5 The adjustment of closed traverses 317 

6.51 Where the start and finish of a traverse are 

fixed 317 

6-52 Traverses which return to their starting point 323 

6.53 Adjusting the lengths without altering the 

bearings 323 

6.54 Adjustment to the length and bearing 330 

6.55 Comparison of methods of adjustment 336 
Exercises 6 (c) (Traverse adjustment) 348 
Exercises 6(d) (General) 352 



XIV 



7 TACHEOMETRY 359 

7.1 Stadia systems — fixed stadia 359 

7.2 Determination of the tacheometric constants 

m and K 360 

7.21 By physical measurement of the instrument 360 

7.22 By field measurement 361 

7.3 Inclined sights 362 

7.31 Staff normal to the line of sight 362 

7.32 Staff vertical 363 

7-4 The effect of errors in stadia tacheometry 367 

7.41 Staff tilted from the normal 367 

7.42 Error in the angle of elevation with the 

staff normal 367 

7.43 Staff tilted from the vertical 368 

7.44 Accuracy of the vertical angle to conform 

to the overall accuracy 371 

7.45 The effect of the stadia intercept assumption 372 
Exercises 7(a) 380 

7.5 Subtense systems 383 

7.51 Tangential method 383 

7.52 Horizontal subtense bar system 388 

7.6 Methods used in the field 392 

7.61 Serial measurement 392 

7.62 Auxiliary base measurement 393 

7.63 Central auxiliary base 395 

7.64 Auxiliary base perpendicularly bisected by 

the traverse line 397 

7.65 Two auxiliary bases 398 
7-66 The auxiliary base used in between two 

traverse lines 400 

Exercises 7(b) 403 

8 DIP AND FAULT PROBLEMS 411 

8.1 Definitions 411 

8.2 Dip problems 413 

8.21 Given the rate and direction of full dip, to find 

the apparent dip in any other direction 413 

8.22 Given the direction of full dip and the rate and 
direction of an apparent dip, to find the rate of 

full dip 413 

8.23 Given the rate and direction of full dip, to find 

the bearing of an apparent dip 415 



XV 

8.24 Given two apparent dips, to find the rate and 

direction of full dip 416 

8.25 Given the rate of full dip and the rate and 
direction of an apparent dip, to find the direction 

of full dip 421 

8.26 Given the levels and relative positions of three 
points in a plane (bed or seam), to find the 

direction and rate of full dip 422 

8.3 Problems in which the inclinations are expressed 

as angles and a graphical solution is required 427 

8.31 Given the inclination and direction of full dip, to find 

the rate of apparent dip in a given direction 427 

8.32 Given the inclination and direction of full dip, 

to find the direction of a given apparent dip 428 

8.33 Given the inclination and direction of two apparent 

dips, to find the inclination and direction of full dip 429 
Exercises 8(a) 429 

8.4 The rate of approach method for convergent lines 432 

8.5 Fault problems 437 

8.51 Definitions 437 

8.52 To find the relationship between the true and 

apparent bearings of a fault 443 

8.53 To find the true bearing of a fault when the throw 

of the fault opposes the dip of the seam 444 

8.54 Given the angle 8 between the full dip of the seam 
and the true bearing of the fault, to find the bearing 

of the line of contact 446 

8.55 To find the true bearing of a fault when the 
downthrow of the fault is in the same general 

direction as the dip of the seam 449 

8.56 Given the angle 8 between the full dip of the 
seam and the true bearing of the fault, to find the 
bearing of the line of contact 449 

8.6 To find the bearing and inclination of the line of 
intersection (AB) of two inclined planes 450 

Exercises 8 (b) (Faults) 452 

Exercises 8 (c) (General) 454 

AREAS 457 

9.1 Areas of regular figures 457 

9.11 Areas bounded by straight lines 457 

9. 12 Areas involving circular curves 459 

9. 13 Areas involving non-circular curves 460 



XVI 

9.14 Surface areas 461 

9.2 Areas of irregular figures 471 

9.21 Equalisation of the boundary to give straight 

lines 471 

9.22 The mean ordinate rule 472 

9.23 The mid-ordinate rule 473 

9.24 The trapezoidal rule 473 

9.25 Simpson's rule 474 

9.26 The planimeter 477 

9.3 Plan areas 481 

9.31 Units of area 481 

9.32 Conversion of planimetric area in square inches 

into acres 482 

9.33 Calculation of area from co-ordinates 482 

9.34 Machine calculations with checks 488 

9.4 Subdivisions of areas 490 

9.41 The subdivision of an area into specified parts 

from a point on the boundary 490 

9.42 The subdivision of an area by a line of known 

bearing 491 

9.43 The sub-division of an area by a line through 

a known point inside the figure 492 

Exercises 9 497 

10 VOLUMES 501 

10. 1 Volumes of regular solids 501 

10.2 Mineral quantities 509 

Exercises 10 (a) (Regular solids) 511 

10.3 Earthwork calculations 513 

10.31 Calculation of volumes from cross-sectional 

areas 513 

Exercises 10 (b) (Cross- sectional areas) 523 

10.32 Alternative formulae for the calculation of 

volumes from the derived cross-sectional areas 525 

10.33 Curvature correction 535 

10.34 Derivation of the eccentricity e of the centroid G 537 

10.4 Calculation of volumes from contour maps 543 

10.5 Calculation of volumes from spot-heights 543 

10.6 Mass-haul diagrams 544 

10.61 Definitions 544 

10.62 Construction of the mass-haul diagram 545 

10.63 Characteristics of the mass-haul diagram 546 



XV11 

10.64 Free-haul and overhaul 546 

Exercises 10 (c) (Earthwork volumes) 552 

11 CIRCULAR CURVES 559 

11.1 Definition 559 

11.2 Through chainage 559 

11.3 Length of curve L 560 

11.4 Geometry of the curve 560 

11.5 Special problems 561 

11.51 To pass a curve tangential to three given 

straights 561 

11.52 To pass a curve through three points 563 
Exercises 11(a) 566 

11.53 To pass a curve through a given point P 567 
Exercises 11(b) (Curves passing through 

a given point) 571 

11.54 Given a curve joining two tangents, to find 
the change required in the radius for an 

assumed change in the tangent length 572 

11.6 Location of tangents and curve 575 

11.7 Setting out of curves 576 

11.71 By linear equipment only 576 

11.72 By linear and angular equipment 580 

11.73 By angular equipment only 580 
Exercises 11(c) 588 

11.8 Compound curves 591 

Exercises 11(d) (Compound curves; 599 

11.9 Reverse curves 600 

Exercises 11(e) (Reverse curves) 605 

12 VERTICAL AND TRANSITION CURVES 607 

12. 1 Vertical curves 607 

12.2 Properties of the simple parabola 608 

12.3 Properties of the vertical curve 609 

12.4 Sight distances 611 

12.41 Sight distances for summits 611 

12.42 Sight distances for valley curves 613 

12.43 Sight distance related to the length of the 

beam of a vehicle's headlamp 615 

12.5 Setting-out data 616 

Exercises 12(a) 624 



XV111 



12.6 Transition curves 627 

12.61 Superelevation 627 

12.62 Cant 628 

12.63 Minimum curvature for standard velocity 628 

12.64 Length of transition 629 

12.65 Radial acceleration 629 

12.7 The ideal transition curve 630 

12.8 The clothoid 632 

12.81 To find Cartesian co-ordinates 632 

12.82 The tangential angle 633 

12.83 Amount of shift 633 

12.9 The Bernouilli lemniscate 634 
12.91 Setting out using the lemniscate 635 

12. 10 The cubic parabola 636 

12.11 The insertion of transition curves 637 

12.12 Setting-out processes 640 

12. 13 Transition curves applied to compound curves 644 

Exercises 12(b) 649 



Abbreviations used for Examination Papers 

E.M.E.U. East Midlands Educational Union 

I.C.E. Institution Of Civil Engineers 

L.U. London University B.Sc. (Civil Engineering) 

L.U./E London University B.Sc. (Estate Management) 

M.Q.B./S Mining Qualifications Board (Mining Surveyors) 

M.Q.B./M Mining Qualifications Board (Colliery Managers) 

M.Q.B./UM Mining Qualifications Board (Colliery Undermanagers) 

N.R.C.T. Nottingham Regional College of Technology 

N.U. Nottingham University 

R.I.C.S./G Royal Institution of Chartered Surveyors (General) 

R.I.C.S./M Royal Institution of Chartered Surveyors (Mining) 

R.I.C.S./ML Royal Institution of Chartered Surveyors (Mining/Land) 

R.I.C.S./Q Royal Institution of Chartered Surveyors (Quantity) 



LINEAR MEASUREMENT 



1.1 The Basic Principles of Surveying 

Fundamental rule 'Always work from the whole to the part*. This 
implies 'precise control surveying' as the first consideration, followed 
by 'subsidiary detail surveying'. 

A point C in a plane may be fixed relative to a given line AB in 
one of the following ways: 

1. Triangulation Angular measurement from a fixed base line. The 
length AB is known. The angles a and /3 are measured. 



ȣ 



B 



a. 



Xe 



li 



.V 



Fig. 1.1(a) 

2. Trilateration Linear measurement only. The lengths AC and BC 
are measured or plotted. The position of C is always fixed provid- 
ed AC + BC > AB. 

Uses: (a) Replacing triangulation with the use of microwave mea- 
suring equipment. 

(b) Chain surveying. 
A 



Bt 




Fig. 1.1(b) 
1 



SURVEYING PROBLEMS AND SOLUTIONS 



3. Polar co-ordinates Linear and angular measurement. 
Uses: (a) Traversing. 

(b) Setting out. 

(c) Plotting by protractor. 



,-° c (s,6) 



BhT 

Fig. 1.1(c) 
4. Rectangular co-ordinates Linear measurement only at right-angles. 
Uses: (a) Offsets. 

(b) Setting out. 

(c) Plotting. 

A 
A 



Bit 



90" 



■OC 



Fig. 1.1(d) 

1.2 General Theory of Measurement 
The following points should be noted: 

(1) There is no such thing as an exact measurement. All measure- 
ments contain some error, the magnitude of the error being dependent 
on the instruments used and the ability of the observer. 

(2) As the true value is never known, the true error is never deter- 



LINEAR MEASUREMENT 3 

mined. 

(3) The degree of accuracy, or its precision, can only be quoted 
as a relative accuracy, i.e. the estimated error is quoted as a fraction 
of the measured quantity. Thus 100 ft measured with an estimated 
error of 1 inch represents a relative accuracy of 1/1200. An error of 
lcm in 100 m = 1/10000. 

(4) Where readings are taken on a graduated scale to the nearest 
subdivision, the maximum error in estimation will be ± l / 2 division. 

(5) Repeated measurement increases the accuracy by y/n, where 

n is the number of repetitions. N.B. This cannot be applied indefinite- 
ly- 

(6) Agreement between repeated measurements does not imply 
accuracy but only consistency. 

1.3 Significant Figures in Measurement and Computation 

If a measurement is recorded as 205 ft to the nearest foot, its most 
probable value is 205 ±0*5 ft, whilst if measured to the nearest 0*1 ft 
its most probable value is 205-0 ± 0-05 ft. Thus the smallest recorded 
digit is subject to a maximum error of half its value. 

In computation, figures are rounded off to the required degree of 
precision, generally by increasing the last significant figure by 1 if the 
following figure is 5 or more. (An alternative is the rounding off with 5 
to the nearest even number.) 

Thus 205-613 becomes 205-61 to 2 places, 
whilst 205-615 becomes 205-62 to 2 places, 

or 205-625 "may also be 205*62, giving a less biased value. 

It is generally better to work to 1 place of decimals more than is 
required in the final answer, and to carry out the rounding-off process 
at the end. 

In multiplication the number of significant figures depends on the 
accuracy of the individual components, e.g., 

if P = x.y, 

then P + 8P = (x + 8x)(y + 8y) 

= xy + x8y + y8x + 8x8y 

Neglecting the last term and substracting P from both sides of the 
equation, 8p = x8y + ySx 

~ P gives S£ = ^§X + y8x = 8y + 8x 



P xy xy 

sp - p (f + t) (11) 



4 SURVEYING PROBLEMS AND SOLUTIONS 

Thus the relative accuracy of the product is the sum of all the 
relative accuracies involved in the product. 

Example 1.1 A rectangle measures 3-82 in. and 7-64 in. with errors of 
± 0*005 in. Express the area to the correct number of significant 
figures. 

P = 3-82 x 7-64 = 29*184 8 in 2 

relative accuracies ° ~ _i_ 

3-82 ~ 750 

0-005 .. 1 



7-64 1500 

500 



SP = 290- + -L-) = ™ 
\750 1500/ 



= ± 0-06 
.-. the area should be given as 29-2in 2 . 

As a general rule the number of significant figures in the product 
should be at least the same as, or preferably have one more significant 
figure than, the least significant factor. 

The area would thus be quoted as 29-18 in 2 

In division the same rule applies. 

Q = - 

y 

Q + 8Q = x + 8x = * + £f - rf^ + ... 
y + 8y y y y 2 

Subtracting Q from both sides and dividing by Q gives 

SQ = Q (?I - *) (1.2) 

Powers R = x n 

R + 8R = (x + 8x) n 

= x n + n8x + ... 

8R n8x . .. , 

— = — — i.e. nx relative accuracy of 

single value. 
8R = n8x (1 . 3) 

Roots This is the opposite relationship 

R = ^x .'. R n = x 

From the above R n + n8R = x + 8x 



LINEAR MEASUREMENT 5 

nSR = Sx 
8R _ 8x_ 
R n ~ nx 

8R = -8x (1.4) 



Example 1.2 If R = (5-01 ± 0-005) 2 

5-01 2 = 25-1001 

8R = 2 x 0-005 = 0-01 
.'. R should be given as 25*10 



Example 1.3 If R = V 25 * 10 ± °* 01 

v'25-10 = 5-009 9 

8R = ^ = 0-005 

.*. R should be given as 5-01 

Example 1.4 A rectangular building has sides approximately 480 
metres and 300 metres. If the area is to be determined to the nearest 
10 m 2 what will be the maximum error permitted in each line, assuming 
equal precision ratios for each length? To what degree of accuracy 
should the lines be measured? 

A = 480 x 300 - 144 000 m 2 

8A = 10 m 2 

8A = _1 = §x Sy 

A 14400 x + y 



but 



8x = 8y . 8x 8y _ 28x 

x y x y ~ x 

8x_ = 1 = 1 

x 2 x 14400 28 800 

i.e. the precision ratio of each line is *„ 

28 800 

This represents a maximum in 480 m of - = 0*016 7 m 

Zq 800 

and in 300 m of -i9p- = 0-0104 m 
2o 800 

If the number of significant figures in the area is 5, i.e. to the 
nearest 10 m 2 , then each line also must be measured to at least 5 sig- 
nificant figures, i.e. 480-00 m and 300-00m. 



SURVEYING PROBLEMS AND SOLUTIONS 



1.4 Chain Surveying 

The chain 

There are two types : 

(a) Gunter's chain 

1 chain* = 100 links = 66 ft 

1 link = 0-66 ft = 7-92 in. 
Its advantage lies in its relationship to the acre 

10 sq chains = 100 000 sq links = 1 acre. 

(b) Engineer's chain 100 links = 100 ft 
(Metric chain 100 links = 20 m 

1 link = 0-2 m) 
Basic figures 

There are many combinations of chain lines all dependent on the 
linear dimensions forming trilateration, Fig. 1 .2. 



Tie line 





C A 



Tie lines 



Fig. 1 .2 Basic figures in chain surveying 




1.41 Corrections to the ground measurements 

Standardisation 

Where the length of the chain or tape does not agree with its nom- 
* See conversion factors, pp. v — vii. 



LINEAR MEASUREMENT 7 

inal value, a correction must be made to the recorded value of a meas- 
ured quantity. 

The following rules apply : 

(1) If the tape is too long, the measurement will be too short — the 
correction will be positive. 

(2) If the tape is too short, the measurement will be too long — the 
correction will be negative. 

If the length of tape of nominal length / is / ± 81, 

fil 
the error per unit length = ± — 

If the measured length is d m and the true length is d t , then 

d t = d m ± d m — 



= ^(l±f) d-5) 



Alternatively, 



1 + 81 actual length of tape 



(1.6) 



d m I nominal length of tape 

d > = *» j 1 ± t) (1 - 5 > 

Example 1.5 A chain of nominal length 100 links, when compared with 
a standard, measures 101 links. If this chain is used to measure a line 
AB and the recorded measurement is 653 links, what is the true 
length AB? 

Error per link = -i— = 0*01 
100 

.'. true length = 653(1 + 0-01) 

= 653 + 6-53 = 659-53 links . 

Alternatively, 

true length = 653 x ^ = 659-53 links . 

Effect of standardisation on areas 

Based on the principle of similar figures, 

true a,ea (.,) . apparent area (,„) x (,££££ 5%. )' 



SURVEYING PROBLEMS AND SOLUTIONS 



or 



A T = A M (l ±y) (1.8) 



Effect of standardisation on volumes 

Based on the principle of similar volumes, 

, / true length of tape V 

true volume V T = apparent volume x ( apparent length of tap J 



ue. 



V r = V„(l ±^)° (110) 

N.B. Where the error in standardisation is small compared to the size 
of the area, the % error in area is approximately 2 x % error in length. 

Example 1.6 A chain is found to be 0*8 link too long and on using it 
an area of 100 acres is computed. 

™. . inn A00-8 \ 2 

The true area = 1UU I - TqTT) 

= 100 x 1-008 2 = 101-61 acres 
alternatively, 

linear error = 0*8% 
••■ area error = 2 x 0*8 = 1*6% 

acreage = 100 + 1*6 acres = 101*6 acres 
This is derived from the binomial expansion of (1 + x) z 

= 1 + 2x + x 2 
i.e .if x is small x z may be neglected 
/. (1 + x) 2 a 1 + 2x 

Correction for slope (Fig. 1.3) 

This may be based on (1) the angle of inclination, (2) the difference 
in level between the ends of the line. 

Fig. 1.3 (page 9) 
Length AC measured (/) 

Horizontal length AB required (h) 

Difference in level between A and C (d) 

Angle of inclination (a) 

Correction to measured length (c) 



LINEAR MEASUREMENT 
h 




Fig. 1.3 
(1) Given the angle of inclination a 
AB = AC cos a 
i.e. h = / cos a (1.11) 

c = I - h 

= I - I cos a 

= /(1-cosa) = / versine a (1-12) 

N.B. The latter equation is a better computation process. 

Example 1.7 If AC = 126-3 m, a = 2°34\ 
byEq.(l.ll) AB = 126-3 cos 2°34' 

= 126-3 x 0-999 = 126-174 m 
or by Eq. (1.12) c = 126-3 (1 - 0-999) 

= 126-3 x 0-001 = 0-126 m 
AB - 126-3 - 0-126 = 126-174 m 

Example 1.8 In chaining, account should be taken of any significant 
effect of the slope of the ground on the accuracy of the horizontal 
length. Calculate the minimum angle of inclination that gives rise to 
relative accuracies of 1/1000 and 1/3000. 

From Eq. (1.12), 

c = I - h = 1(1 - cos a) 

c = _J_ 
T 1000 



If 



1 - cos a 



10 SURVEYING PROBLEMS AND SOLUTIONS 

cos a = 1 - 0-001 = 0-999 
a = 2°34' (i.e. 1 in 22) 



Also, if -j = 


= = 1 - cos a 

3000 




cos a = 1 - 0-00033 




= 0-99967 




a = 1°29' (i.e. 1 in 39) 


If the difference in level 


, d, is known 


h = 


(I 2 - d 2 y = j(/-d)x (/+ d)}* 


or I 2 = 


h 2 + d 2 


= 


(/ - cf + d 2 


= 


I 2 - 2lc + c 2 + d 2 


.-. c 2 - 2lc = 


-d 2 


c(c-2l) = 


-d 2 


c = 


-d 2 
c-2l 


c ~ 


d z 

— as c is small compared 



(1.13) 



Rigorously, using the binomial expansion, 
c - I - (I 2 - d 2 y 

-'-<-£)' 

d 2 d' 
= Tl ~ gji ♦ ••• (1.15) 

The use of the first term only gives the following relative accura- 
cies (the units may be ft or metres). 

Gradient Error per 100 ft or m Relative accuracy 

lin4 0-051 ft (or m) 1/2000 

1 in 8 0-003 1 ft (or m) 1/30 000 

1 in 10 0-001 3 ft (or m) 1/80000 

1 in 20 0-000 1 ft (or m) 1/1 000 000 

Thus the approximation is acceptable for: 

Chain surveying under all general conditions. 

Traversing, gradients up to 1 in 10. 

Precise measurement (e.g. base lines), gradients up to 1 in 20. 



LINEAR MEASUREMENT 11 

For setting out purposes 

Here the horizontal length (h) is given and the slope length (/) is 
required. 

/ = h sec a 

c = h sec a - h 

= h(sec a - 1) (1.16) 

Writing sec a as a series 1 + ^- + ^+ •••, where a is in radians, 
i „ x see p. 72. 

ho. 2 

— -s- («• in radians) (1.17) 

~ ^(0-017 45a) 2 

2i 1-53 ft x 10~ 4 x a 2 (a in degrees) (1.18) 

- 1*53 x 10" 2 x a 2 per 100 ft (or m) (1.19) 

Example 1.9 If h = 100 ft (orm), a = 5°," 
by Eq. (1.16) c = 100(1-003 820-1) 

= 0-382 Oft (orm) per 100 ft (orm) 
or by Eq. (1.18) c = 1-53 x 100 x 10" 4 x 5 2 

= 1-53 x 25 x 10~ 2 

= 0-382 5 ft (or m) per 100 ft (or m) 
Correction per 100ft (orm) 

1° 0-015 ft (orm) 6° 0-551 ft (orm) 

2° 0-061 ft (orm) 7° 0-751 ft (orm) 

3° 0-137 ft (orm) 8° 0-983 ft (orm) 

4° 0-244 ft (orm) 9° 1-247 ft (orm) 

5° 0-382 ft (orm) 10° 1-543 ft (orm) 
If the difference in level, d, is given, 

I 2 = h 2 + d 2 

(h + cf = h 2 + d 2 

h 2 + 2hc + c 2 = h 2 + d 2 
c(2h + c) = d 2 

c- -*- 
2h + c 



12 



SURVEYING PROBLEMS AND SOLUTIONS 



2h 



or rigorously 



2h + 8h 3 



(1.20) 



(1.21) 



N.B. If the gradient of the ground is known as 1 vertical to n horizon- 

tal the angle of inclination a ~ S (1 rad ~ 57" 3°) 

n 



e.g. 1 in 10 gives 



57 
10 



5-7 c 



To find the horizontal length h given the gradient 1 in n and the 
measured length I 



n\fn z + 1 



/ 


V« 2 + i 


n 2 + 1 


h -. 


lny/n z + 1 
« 2 +l 


(1.22) 




As an alternative to the above, 

h = I - c 



- '"IT 
but if the gradient is given as 1 in n, then 

d c* — 
n 



Fig. 1.4 



h ^ I 



2nH 



\ 2n 2 J 
This is only applicable where n > 20. 



(1.23) 



Example 1.10 If a length of 300 ft (orm) is measured on a slope of 1 
in 3, the horizontal length is given as: 

by Eq. (1.22) h = 300x3^10 = 9Q x 3 . 1623 

10 

- 284-61 ft (orm) 



LINEAR MEASUREMENT 

To find the inclined length I given the horizontal length h and the 
gradient (1 in n) 



13 



± = V" 2 + 1 
h n 

, __ hy/n z + 1 



(1-24) 



Example 1.11 If h = 300 ft (orra) and the gradient is 1 in 6, 

by Eq. (1.24) / = 3Q0 ^ = 50 x 6-083 

6 

= 304 -15 ft (or m) 

1.42 The maximum length of offsets from chain lines 

A point P is measured from a chain line ABC in such a way that 
B, P is measured instead of BP, due to an error a in estimating the 
perpendicular, Fig. 1.5. 




Fig. 1.5 
On plotting, P, is fixed from B, . 

Thus the displacement on the plan due to the error in direction a 
PP, = B,P a(radians) 

_ /a, 
= 206265 

(N.B. 1 radian = 206265 seconds of arc) 



14 SURVEYING PROBLEMS AND SOLUTIONS 

If the maximum length PP % represents the minimum plotable point, 

12 



i.e. 0*01 in which represents ^— xft, where x is the representative 



fraction 1/x, then 

0-000 83 x = la 



I 



206 265 
171-82 x 



a 

Assuming the maximum error a = 4°, i.e. 14400 , 

= 171-82* o-012 x (1.25) 

14400 

If the scale is 1/2500, then x = 2500, and 

/ = 2500 x 0-012 = 30ft (^10 m) 

If the point P lies on a fence approximately parallel to ABC, 
Fig. 1.6, then the plotted point will be in error by an amount P^P 2 
= 1(1- cos a). (Fig. 1.5). 

Boundary line 



12 (1 - cos a) 

Example 1.12 If a = 4°, by Eq. 1.26 

0-01 x 



fl C 

Fig. 1 .6 
, = 0-01 x (1>26) 



/ 



12 x (1-0-9976) 

0-35 x (1.27) 



Thus, if x= 2500, 

/ = 875 ft (267 m) 
The error due to this source is almost negligible and the offset is only 
limited by practical considerations, e.g. the length of the tape. 

It is thus apparent that in fixing the position of a point that is 
critical, e.g. the corner of a building, the length of a perpendicular off- 
set is limited to 0*012 x ft, and beyond this length tie lines are required, 



LINEAR MEASUREMENT 

the direction of the measurement being ignored, Fig. 1.7. 



15 




Fig. 1.7 
1.43 Setting out a right angle by chain 

From a point on the chain line (Fig. 1.8) 
(a) (i) Measure off BA = BC 

(ii) From A and C measure off AD = CD 
(Proof: triangles ADB and DCB are congruent, thus ABD = DBC = 90 c 
as ABC is a straight line) 



Fig. 1.8 



Z 



/ 



/ 



/ 



7k 



\ 



\ 




\ 



8 



B 
Fig. 1.9 



(b) Using the principle of Pythagoras, 

z z = x z + y 2 (Fig. 1.9) 
By choosing suitable values the right angle may be set out. 
The basic relationship is 

x:y:z :: 2n + l : 2n(rc + l) : 2n(n + l)+l. (1.28) 

If n = I, 2n + l=3 

2n(n + 1) = 4 
2n(n + 1) + 1 = 5. 
Check: \2n(n + 1) + l} 2 = (2n 2 + 2n + l) 2 

(2n + if + \2n(n + 1)| 2 = 4n 2 + 4n + 1 + 4n 4 + 8n 3 + 4n 2 
= 4n z + 8n 3 + 8n 2 + 4n + 1 



16 



SURVEYING PROBLEMS AND SOLUTIONS 





= (2n 2 + 2n + 


If. 




Check: 


5 2 - 3 2 + 4 2 






or 


25 - 9 + 16. 






Similarly, if n = 3/4, 








In + 1 = 


6 , 10 
4 4 


= 


40 
16 


2n(rc + 1) = 


6/3 , \ 6 7 
4U +1 j = 4 X 4 = 


42 
16 


2n(n + 1) + 1 = 


16 


16 
16 


58 
16 



Thus the ratios become 40 : 42 
: 58 and this is probably the best 
combination for 100 unit measuring 
equipment; e.g. on the line ABC, 
Fig. 1.10, set out BC = 40 units. 
Then holding the ends of the chain 
at B and C the position of D is 
fixed by pulling taut at the 42/58 
on the chain. 



Alternative values for n give the following: 

n = 2 5, 12, 13 (Probably the best ratio for 30 m tapes) 

n = 3 7, 24, 25 

n = 4 9, 40, 41. 

1.44 To find the point on the chain line which produces a perpendic- 
ular from a point outside the line 




(1) When the point is accessible 
(Fig. 1.11). From the point D swing 
the chain of length > DB to cut the 
chain line at a and b. The required 
position B is then the mid-point of 
ab. 




LINEAR MEASUREMENT 



17 



(2) When the point is not accessible 
(Fig. 1.12). From D set out lines 
Da and Db and, from these lines, 
perpendicular ad and be. The inter- 
section of these lines at X gives the 
the line DX which when produced 
gives B, the required point. 




To set out a line through a given 
point parallel to the given chain 
line (Fig. 1.13). Given the chain 
line AB and the given point C. 
From the given point C bisect the 
line CB at X. Measure AX and 
produce the line to D such that AX 
= XD. CD will then be parallel to AB. 




1.45 Obstacles in chain surveying 




18 



SURVEYING PROBLEMS AND SOLUTIONS 



(1) Obstacles to ranging 

(a) Visibility from intermediates (Fig. 1.14). Required to line C and 
D on the line AB. 

Place ranging pole at d, and line in c, on line Ad } . From B ob- 
serve c, and move d 2 on to line Be, . Repetition will produce c 2 , c 3 and 
d 2 , d 3 etc until C and D lie on the line AB. 

(b) Non-visibility from intermediates (Fig. 1.15). 
Required to measure a long line AB 

in which A and B are not inter- 
visible and intermediates on these 
lines are not possible. 

Set out a 'random line' AC 
approximately on the line AB. 

From B find the perpendicular 
BC to line AC as above. Measure 
AC and BC. Calculate AB. 

(2) Obstacles to chaining 
(a) No obstacle to ranging 

(i) Obstacle can be chained around. There are many possible 
variations depending on whether a right angle is set out or not. 



o*= 

A 




B 



Fig. 1.15 






A6 (m) 



Fig. 1.16 By setting out right angles 

I Set out equal perpendiculars Bb and Cc; then be - BC. 

II Set out Bb. Measure Bb and bC. Compute BC. 

HI Set out line Bb. At b set out the right angle to give C on the 
chain line. Measure Bb and bC. Compute BC. 

IV and V Set out parallel lines be as described above to give 
similar figures, triangles BCX and bcX. 



Then 



BC = 



be x BX 

bX 



(1.29) 



LINEAR MEASUREMENT 



19 




Fig. 1.17 
VI Set out line be so that bB = Be. Compute BC thus, 

BC* = CO 2 Be + jCcf bB _ 
be 

but bB = Be, 

. RC z BbjbC 2 - Cc 2 ) _. DK 
- ^ = 2fib -BbxBb 

= UbC z + Cc 2 ) - Bb 2 



(1.30) 



(1.31) 



Proof. 

In Fig. 1.18 using the cosine rule (assuming 6 > 90°), see p. 81 



,2 _ v 2 



x 2 + d 2 + 2xd cos0 and 
? 2 = y 2 + d 2 - 2yd cosO 



2d cos 6 = 



p - JT 



_ y2 +d 2_ ? 2 



P 2 y - * 2 y - d 2 y = xy 2 + d 2 x - q 2 x 

d 2 (x + y) = <? 2 x + p 2 y - xy(x + y) 



20 



SURVEYING PROBLEMS AND SOLUTIONS 



d 2 = 



q z x + p 2 y 



x + y 



- xy 



--/ 



Q x + p y 
x + y 



- xy (1.32) 



If x = y, 



(ii) Obstacle cannot be chained 
around. A river or stream represents 
. this type of obstacle. Again there 
are many variations depending on 
whether a right angle is set out or 
not. 



By setting out right angles (Fig. 1.19). 

A random line DA^ is set out and from perpendiculars at C and B 
points C and B are obtained. 

By similar triangles DC,C and C 1 B 1 B 2 , 




DC 
CB 

DC 



CC X 
Bfi, - CC y 
CB x CC, 



BB. - CC, 




Fig. 1.20 



LINEAR MEASUREMENT 



21 



Without setting out a right angle (Fig. 1.20). 

A point F is chosen. From points B and C on line AE, BF and 
CF are measured and produced to G and H. BF = FG and CF = FH. 
The intersection of DF and GH produce to intersect at J. Then 
HJ = CD. 

(iii) Obstacles which obstruct ranging and chaining. The obstruction, 
e.g. a building, prevents the line from being ranged and thus produced 
beyond the obstacle. 

By setting out right angles (Fig. 1 .21) 

On line ABC right angles are set out at B and C to produce B 
and C, , where SB, = CC, . ' 

B, C, is now produced to give D, and E t where right angles are 
set out to give D and E, where D^D=E^E=BB y = CC, . D and E are 
thus on the line ABC produced and 0,0, = DC. 





A A 

Fi B-l.21 Fig. 1.22 

Without setting out right angles (Fig. 1.22) 

On line ABC, CB is measured and G set out to form an equilateral 
triangle, i.e. CB= CG = BG. BG is produced to J. 

An equilateral triangle HKJ sets out the line JE such that JE = 
BJ. 

A further equilateral triangle ELD will restore the line ABC pro- 
duced . 

The missing length BE = BJ = EJ . 



22 SURVEYING PROBLEMS AND SOLUTIONS 

Exercises 1 (a) 

1. The following measurements were made on inclined ground. 
Reduce the slope distances to the horizontal giving the answer in feet. 

(a) 200-1 yd at 1 in 2^ 

(b) 485*5 links at 1 in 5-75 

(c) 1/24 th of a mile at 1 in 10-25 

(Ans. (a) 557-4 ft (b) 315-7 ft (c) 218-9 ft) 

2. Calculate the acreage of an area of 4 in 2 on each of the plans 
drawn to scale, 2 chains to 1 in., 1/63 360, 1/2500 and 6 in. to 1 mile 
respectively. 

(Ans. 1-6, 2560, 3-986, 71-1 acres) 

3. A field was measured with a chain 0-3 of a link too long. The 
area thus found was 30 acres. What is the true area? 

(I.C.E. Ans. 30-18 acres) 

4. State in acres and decimals thereof the area of an enclosure mea- 
suring 4 in. square on each of three plans drawn to scale of 1/1584, 
1/2500, 1/10560 respectively. 

(Ans. 6-4, 15-9, 284-4 acres) 

5. A survey line was measured on sloping ground and recorded as 
386-6 ft (117-84 m). The difference of elevation between the ends was 
19-3 ft (5-88 m). 

The tape used was later found to be 100-6 ft (30-66 m) when com- 
pared with a standard of 100 ft (30-48 m). 

Calculate the corrected horizontal length of the line. 

(Ans. 388-4 ft (118-38 m)) 

6. A plot of land in the form of a rectangle in which the length is 
twice the width has an area of 180000 ft 2 . 

Calculate the length of the sides as drawn on plans of the follow- 
ing scales. 

(a) 2 chains to 1 inch, (b) 1/25000. (c) 6 inch to 1 mile. 
(Ans. (a) 4-55 x 2-27 in. (b) 0-29 x 0-14 in. (c) 0-68 x 0-34 in.) 

7. (a) Express the following gradients in degrees to the horizontal: 
1 in 3, 1 in 200, 1 in 0-5, being vertical to horizontal in each case. 

(b) Express the following scales as fractions: 6 in. to 1 mile, 1 in. 
to 1 mile, 1 in. to 1 chain, 1/8 in. to 1 ft. 

(c) Express the following scales as inches to 1 mile: 1/2500, 
1/500, 1/1080. 

(M.Q.B./UM Ans. (a) 18°26', 0°17', 63°26' 

(b) 1/10560, 1/63360, 1/792, 1/96 

(c) 25-34, 126-72, 58-67) 



LINEAR MEASUREMENT 23 

8. Find, without using tables, the horizontal length in feet of a line 
recorded as 247*4 links when measured 

(a) On ground sloping 1 in 4, 

(b) on ground sloping at 18°26' (tanl8°26' = 0-333). 

(Ans. (a) 158-40 (b) 154-89 ft) 

9. Show that for small angles of slope the difference between the 
horizontal and sloping lengths is h z /2l (where h is the difference of 
vertical height of the two ends of a line of sloping length /). 

If errors in chaining are not to exceed 1 part in 1000, what is the 
greatest slope that can be ignored? 

(L.U./E Ans. 1 in 22*4) 

1.5 Corrections to be Applied to Measured Lengths 

For every linear measurement the following corrections must be 
considered, the need for their application depending on the accuracy 
required. 

1. In all cases 

(a) Standardisation. (b) Slope. 

2. For relative accuracies of 1/5000 plus 

(a) Temperature. (b) Tension, 

(c) Sag. (where applicable) 

3. For special cases, 1/50000 plus 

(a) Reduction to mean sea level. (b) Reduction to grid. 

Consideration has already been given, p. 6/9, to both standardisa- 
tion and reduction to the horizontal as they apply to chain surveying 
but more care must be exercised in precise measurement reduction. 

1.51 Standardisation 

The measuring band in the form of a tape or wire must be compared 
with a standard under specified conditions of temperature (t s ) and ten- 
sion (T s ) . If there is any variation from the nominal length then a stan- 
dardisation correction is needed as already shown. A combination of 
temperature and standardisation can be seen under correction for tem- 
perature. 

1.52 Correction for slope 

Where the inclination of the measured length is obtained by mea- 
surement of the vertical angle the following modification should be 
noted. 

Let the height of the instrument be ft, 
the height of the target h 2 



24 



SURVEYING PROBLEMS AND SOLUTIONS 



the measured vertical angle 6 
the slope of the measured line a 
the length of the measured line / 




Fig. 1.23 

In Fig. 1.23, a = d + 8$. 

In triangle A A B Z B^ by the sine rule (see p. 80), 

(h, -h z ) sin (90 + 0) 



sin 86 



86" 



I 

(fc, - h 2 ) cos 6 
/ 

206 265 (h } -h z ) cos 6 
_ 



(1.34) 
(1.35) 



N.B. 



The sign of the correction conforms precisely to the equation. 

(1) If /i i = h z , 86 = a = 6 

(2) If h\ < h Z z and 6 is +ve, 86 is -ve (Fig. 1.24a) 

(3) If h,> h z and 6 is -ve, 86 is -ve (Fig. 1.24d) 

if a is +ve, 86 is +ve (Fig. 1.24b) 

(4) If /i, < h z and 6 is -ve, 86 is +ve (Fig. 1.24c) 



Example 1.13 

If ^ = 4-5ft(l-37m), h z = 5-5ft (1.68m), 6 = +4°30' 
/ = 350 ft (106-68 m) 

206 265 (4-5 - 5-5) cos4°30' 



then 86 = 



350 



- -588" 
= -0°09'48" 
a = +4 o 30'00" - 0°09'48" 
= +4° 20 '12" 



LINEAR MEASUREMENT 



25 



/> 4 -/>, 



(a) 



(b) 




(c) 




(d) 



Fig. 1.24 
Correction to measured length (by Eq. 1.12), 
c = -1(1 -cos0°) 

= -350(1 -cos 4° 20' 12") 
= -350(1-0-99714) 

= -350 x 0-00286 = - 1-001 ft (0-3051 m) 
••• Horizontal length = 348-999 ft (106-3749 m) 
If the effect was ignored; 

Horizontal length = 350 cos 4° 30' 

= 348-922 ft (106-3514 m) 
••• Error = 0-077 ft (0-0235 m) 



26 SURVEYING PROBLEMS AND SOLUTIONS 

1.53 Correction for temperature 

The measuring band is standardised at a given temperature (t 8 ). 
If in the field the temperature of the band is recorded as (* m ) then the 
band will expand or contract and a correction to the measured length 
is given as 

c = la(t m - t a ) (1.36) 

where / = the measured length 

a = the coefficient of linear expansion of the band metal. 

The coefficient of linear expansion (a) of a solid is defined as 
'the increase in length per unit length of the solid when its temper- 
ature changes by one degree'. 

For steel the average value of a is given as 

6-2 x 10" 6 per °F 

Since a change of 1 °F = a change of 5/9 °C, using the value above 

gives 

a = 6-2 x 10" 6 per 5/9 °C 

= 11-2 x 10" 6 per °C 

The range of linear coefficients a is thus given as : 



Steel 
Invar 



per 1°C 



-6 



10-6 to 12-2 (x 10 ) 
5-4 to 7-2 (xl0~ 7 ) 



per 1°F 
5-9 to 6-8 
3 to 4 

To find the new standard temperature t' 8 which will produce the 
nominal length of the band. 

Standard length at t a = I ± 81 

To reduce the length by 5/: 

81 = (/ ± 5/).a.r 

where t = number of degrees of temperature 

change required 

81 



t = 



(I ±8t)a 



' ; - »• * orkz (137 > 



As 81 is small compared with /, for practical purposes 

81 
la 



K = t s ± £f (1.38) 



LINEAR MEASUREMENT 27 

Example 1.14 A traverse line is 500 ft (152*4 m) long. If the tape 
used in the field is 100 ft (30-48 ra) when standardised at 63 °F (17-2 °C), 
what correction must be applied if the temperature at the time of mea- 
surement is 73 °F (22-8 °C)? 

(Assume a = 6-2 x 10~ 6 per deg F 
= 11-2 x 10~ 6 per deg C) 
From Eq. (1.36) 

c m = 500 x 6-2 x 10~ 6 x (73 - 63) 
= +0-031 Oft 



or 



c (m) = 152-4 x 11-2 x 10" 6 x (22-8 - 17-2) 
= +0-009 6 m 



Example 1.15 If a field tape when standardised at 63 °F measures 
100-005 2 ft, at what temperature will it be exactly the nominal value? 

(Assume a = 6-5 x 10" 6 per deg F) 
SI = +0-0052 ft 

•'. from Eq. (1.37) t' =63 0-0052 

100 x 6-5 x 10" 6 

= 63°F-8°F 

= 55 °F 

In its metric form the above problem becomes: If a field tape when 
standardised at 17-2 °C measures 100-005 2 m, at what temperature will 
it be exactly the nominal value? 

(Assume a = 11-2 x 10" 6 per deg C) 

81 = +0-005 2 m 

.'. from Eq.(1.37) t' s = 17#2 _ 0-0052 

100 x 11-2 x 10" 6 

= 17-2 °C - 4-6 °C 

= 12-6 °C (=54-7°F) 

1.54 Correction for tension 

The measuring band is standardised at a given tension (T s ). If in 
the field the applied tension is (T m ) then the tape will, due to its own 
elasticity, expand or contract in accordance with Hooke's Law. 

A correction factor is thus given as 

L(T m - T s ) 
c = A _ E (1.39) 



28 SURVEYING PROBLEMS AND SOLUTIONS 

where L = the measured length (the value of c is in the same unit 
as L), 

A = cross-sectional area of the tape, 
E = Young's modulus of elasticity i.e. stress/strain. 
The units used for T, A and E must be compatible, e.g. 
T(lbf) A (in 2 ) E (lbf/in 2 ) 

or T,(kgf) 4, (cm 2 ) E, (kgf/cm 2 ) (metric) 

or T 2 (N) A,(m 2 ) E 2 (N/m 2 ) (new S.I. units) 

Conversion factors 

lib = 0-453592 kg 
1 in 2 = 6-451 6 x 10~ 4 m 2 
.'. lib/in 2 = 703-070 kg/m 2 

Based on the proposed use of the International System of Units (S.I. 
units) the unit of force is the Newton (N), i.e. the force required to 
accelerate a mass of 1 kg 1 metre per second per second . 

The force 1 lbf = mass x gravitational acceleration 

= 0-453592 x 9-80665 m/s 2 (assuming standard value) 
= 4-448 22N 
lkgf = 9-806 65 N (1 kg = 2-204 62 lb) 
whilst for stress 1 lbf/in 2 = 6894-76 N/m 2 
For steel, E ~ 28 to 30 x 10 6 lbf/in 2 (British units) 

~ 20 to 22 x 10 5 kgf/cm 2 (Metric units) 

ot 19-3 to 20-7 x 10 10 N/m 2 (S.I. units) 

For invar, E ~ 20 to 22 x 10 6 lbf/in 2 

^ 14 to 15-5 x 10 5 kgf/cm 2 

~ 13-8 to 15*2 x 10 ,0 N/m 2 
N.B. (1) If T m = T s no correction is necessary. 

(2) It is generally considered good practice to over tension to 
minimise deformation of the tape, the amount of tension 
being strictly recorded and the correction applied. 

(3) The cross-sectional area of the tape may be physically 
measured using a mechanical micrometer, or it may be com- 
puted from the total weight W of the tape of length L and 
a value p for the density of the material. 

W 
A - j- (1.40) 



LINEAR MEASUREMENT 29 

Example 1.16 A tape is 100 ft at a standard tension of 251bf and mea- 
sures in cross-section 0-125 in. x 0-05 in. If the applied tension is 
20 lbf and E = 30 x 10 5 lbf/in 2 , calculate the correction to be applied. 

Rv F„1C.Q . 10 ° X (20 - 25) 

By t-q. l.iy c = . i = -0*009 7 ft 

(0-125 x 0-05) x (30 x 10 6 ) ° °° 27 " 

Converting the above units to the metric equivalents gives 

c = 30-48 m x (9-072 - 11-340) kgf 

(40-32 x 10~ 7 )m 2 x (21-09 x 10 9 ) kgf/m 2 
= -0-008 13 m (i.e. -0-002 7 ft) 
Based on the International System of Units, 

2-268 kgf = 2-268 x 9-806 65 N = 22-241 N 
or 5 lbf = 5 x 4-448 22 N = 22-241 N. 

For stress, 

(21-09 x 10 10 ) kgf/m 2 = 21-09 x 10 9 x 9-80665 = 20-684 x 10 ,o N/m 2 
or 

(30 x 10 6 ) lbf/in 2 = 30 x 10 6 x 6894-76 = 20-684 x 10'° N/m 2 
Thus, in S.I. units, 

30-48 x 22-241 N 



c = 



(40-32 x 10 -7 ) x (20-684 x 10 ,o )N/m 2 
= - 0-008 13 m 

Measurement in the vertical plane 

Where a metal tape is freely suspended it will elongate due to the 
applied tension produced by its own weight. 

The tension is not uniform and the stress varies along its length. 

Given an unstretched tape AB and a stretched tape AB, , Fig. 1.25, 
let P and Q be two close points on the tape which become ^Q, under ' 
tension. 

If AP = x, AP t = x + s, where s is the amount of elongation of AP . 
Let PQ = dx, then P, Q, = dx + ds and the strain in %Q,= — as ds 
is the increase in length and dx is the original length. 

If T is the tension at P, , in a tape of cross-section A, and E is 
Young's modulus, then 

T = EA*1 m 

dx (!) 

Given that the load per unit length at P, is w then in P, Q, the load 
= w.dx being the difference in tension between P, and Q, 



30 



SURVEYING PROBLEMS AND SOLUTIONS 



if the tension at Q is T + dT 

T-(T + dT) = wdx 
i.e. dT = -wdx 



(2) 



x + s 



dx+ds 




Fig. 1.2 5 Elongation in a suspended tape 

In practice the value w is a function of x and by integrating the 
two equations the tension and extension are derived. 

Assuming the weight per unit length of the tape is w with a sus- 
pended weight W, then from (2) 

dT = -wdx 

T = -wx + c (3) 



and (1) 



T = EA^. 
dx 

ds 



EA -t— = -wx + c 
dx 



EAs = 



iwx 2 + ex + d 



(4) 



When T = W, x= I and when x = 0, s = 

W = - wl + c i.e. c = W + wl 

and d = 0. 



LINEAR MEASUREMENT 31 

Therefore putting constants into equations (3) and (4) gives 

T = -wx + W + wl 

T = W + w(l-x). (i.4i) 

and EAs = - 1 wx 2 + Wx + wlx 



If x = /, then 



and if W = 0, 



= Wx + ±w(2lx- x 2 ) 

= ml Wx + i w & lx -* 2 )] d-42) 

S= A[^ + I w/2 ] (1-45) 



S ~ 2E4 



(1.44) 



Example 1.17 Calculate the elongation at (1) 1000 ft and (2) 3000 ft 
of a 3000 ft mine-shaft measuring tape hanging vertically due to its 
own weight. 

The modulus of elasticity is 30 x 10 6 lbf/in 2 ; the weight of the 
tape is 0*05 lbf/ft and the cross-sectional area of the tape is 0*015 in 2 . 
From Eq.(1.42) 

s = ljz[ w *+ i"(2lx-x z )] 
As W = 0, 

s= m [2lx - x * ] 

when x = 1000 ft 
/ = 3000 ft 

0-05 
S = 2 x 30 x 10* x 0-015 C2 X 300 ° X 100 ° " 100(f] 
~ °- 05 ^xl0« 



2 x 30 x 10° x 0-015 
when x = / = 3000 ft. 

■2 



From Eq. (1.44) s = — ^L 
2EA 



0-05x3000 2 „„„„, 

= 0-500 ft 



2 x 30 x 10 6 x 0-015 



32 SURVEYING PROBLEMS AND SOLUTIONS 

Example 1.18 If the same tape is standardised as 3000 ft at 451bf 
tension what is the true length of the shaft recorded at 2998*632 ft? 

I„Eq.(1.44) s- -S? -i™: 

' 2EA EA 

i.e. T = \W 

where W = total weight of tape = 3000 x 0-05 = 150 Ibf 

Applying the tension correction, Eq.(1.39), 

L(J m - T s ) 
C = EA 

3000(75-45) = 30 x 10 2 x 30 = . 200ft 
30 x 10 6 x 0-015 30 x 10 6 x 0-015 

.'. true length 

= 2998-632 + 0-2 = 2998-832 ft 

1.55 Correction for sag 

The measuring band may be standardised in two ways, (a) on the 
flat or (b) in catenary. 

If the band is used in a manner contrary to the standard conditions 
some correction is necessary. 

(1) // standardised on the flat and used in catenary the general 
equation for correction is applied, viz. 

w 2 / 3 



c = - r-^ d-45) 

(1.46) 



24 T 2 

W 2 l 



24 T 

where w = weight of tape or wire per unit length 

W = wl = total weight of tape in use, 

T = applied tension. 
N.B. the units w and T must be compatible (lbf, kgf or N) 




Measuring 

head Catenary curve 

Fig. 1.26 Measurement in catenary 

(2) // standardised in catenary 

(a) The length of the chord may be given relative to the length 



LINEAR MEASUREMENT 33 

of the tape or 
(b) the length of the tape in catenary may be given. 

(i) If the tape is used on the flat a positive sag correction must be 
applied 

(ii) If the tape is used in catenary at a tension T m which is different 
from the standard tension T s , the correction will be the difference 
between the two relative corrections, i.e. 

W z l r 1 IT 

C --24[n;-Tl\ (1-47) 

If T m > Ts the correction will be positive. 

(iii) If standardised in catenary using a length l a and then applied in 
the field at a different length l m , the correction to be applied is 
given as 



c = ?i 



L. //, 3 w 2 HW' 



h \24I 2 24T 

Alternatively, the equivalent tape length on the flat may be com- 
puted for each length and the subsequent catenary correction applied 
for the new supported condition, i.e. if /, is the standard length in 
catenary, the equivalent length on the ground = /. + c„ where c = the 
catenary correction. s 

If l m is the applied field length, then its equivalent length on the 

flat = W + c s ) 

Applying the catenary correction to this length gives 

lm + C = £ (l s + C j _ Cm 



= L +^ 



Thus the required correction 



'■m^s 
i — Cm 

I /'> 2 \ l 3 w 2 



l s \24T Z I 241' 

j-^yi ( Z * ~ l D as Eq. (1 .48) above . 



34 



SURVEYING PROBLEMS AND SOLUTIONS 



The sag correction is an acceptable approximation based on the 
assumption that the measuring heads are at the same level. If the heads 
are at considerably different levels, Fig. 1.27, the correction should be 



c = c, cos 2 



(l±^sin*) 



(1.49) 



the sign depending on whether the tension is applied at the upper or 
lower end of the tape. 




Measuring 
head 



Fig. 1.27 
For general purposes c = c, cos 2 

w 2 Pcos 2 g 
24 T 2 



(1.50) 



The weight of the tape determined in the field 

The catenary sag of the tape can be used to determine the weight of 
of the tape, Fig. 1.28 




Fig. 1.28 Weight of the tape determined in the field 

If y is the measured sag at the mid-point, then the weight per unit 
length is given as 

STy 



or the amount of sag 



w = 



y = 



i 2 

wl z 
ST 



(151) 



(1.52) 



where w = weight/unit length, y = vertical sag at the mid-point, 

T = applied tension, / = length of tape between supports. 



LINEAR MEASUREMENT 35 

Example 1.19 Calculate the horizontal length between two supports, 
approximately level, if the recorded length is 100.237 ft, the tape 
weighs 15 ozf and the applied tension is 20 lbf . 

From Eq.(1.46) c = - W * 1 

24T 2 

The value of / is assumed to be 100 for ease of computation. 
Then 



5\ 2 
jjx 100 

24 x 20 2 



= -0-0092 ft 
True length = 100-2370 - 0-0092 

= 100-227 8 ft 

Example 1.20 A 100 ft tape standardised in catenary at 25 lbf is used 
in the field with a tension of 20 lbf. Calculate the sag correction if 
w = 0-021 lbf/ft. 



From Eq. (1.47) c = -[1^1 



T 2 T 2 ) 



- 100 3 x 0-021 2 / 1 _1_ 
24 (20 5 "25 1 

= -0-01656 i.e. -0-016 6 ft. 

Example 1.21 A tape 100 ft long is suspended in catenary with a ten- 
sion of 30 lbf. At the mid-point the sag is measured as 0-55 ft. Cal- 
culate the weight per ft of the tape. 
From Eq. (1.51), 

8Ty 8 x 30 x 0-55 

T - 10000 = 0-0132 IWft. 



Based on S.I. units these problems become 
1- 19(a) Calculate the horizontal length between two supports approx- 
imately level if the recorded length is 30-552 2 m; the tape weighs 
0-425 kgf and the applied tension is 9-072 kgf. 

Converting the weight and tension into units of force, 
30-5522 (0-425 x 9-806 65) 2 



JV VJA Z, \VftZ.D X. 3'OUD 03 V 

24 (9-072 x 9-806 65) 2 



36 SURVEYING PROBLEMS AND SOLUTIONS 

Thus there is no significance in changing the weight W and tension T 
into units of force, though the unit of tension must be the newton. 

30-552 2 / 0-425 \ 2 
24 \ 9-072/ 
= -0-002 8 m (-0-009 2 ft). 

1.20(a) A 30*48 m tape standardised in catenary at 111-21 N is used 
in the field with a tension of 88-96 N. Calculate the sag correction if 
w= 0-0312kgf/m. 

Conversion of the mass/ unit length w into a total force gives 

30-48 x 0-0312 x 9-80665 = 9-326 N. 



Eq. (1 .47) becomes 



c = 



24 



'-- -) 
T 2 T z 

\88-96 2 111-21 2 / 



= -30-48 x 9-326 
24 

= -0-00504 m (-0-016 6 ft). 

1.21(a) A tape 30-48 m long is suspended in catenary with a tension 
of 133-446 N. At the mid-point the sag is measured as 0-168 m. Cal- 
culate the weight per metre of the tape. 

Eq. (1.51) becomes 

n (/ , 8 x T x y 0-816 Ty 

w (kgf/m) = - = — 

9-80665 / 2 I 2 

0-816 x 133-446 x 0-168 

30-48 2 

= 0-019 6 kgf/m (0-013 2 lbf/ft) 

Example 1.22 A tape nominally 100 ft is standardised in catenary at 
lOlbf and is found to be 99-933 ft. If the weight per foot is 0-01 lbf, 
calculate the true length of a span recorded as 49*964 ft. 

Standardised length = 99-933 ft 

Sag correction for 100 ft 

Q-01 2 x 100 3 rtftyIoft . 

c i = — 7Ta T^T - = 0-042 ft 

1 24 x 10 

True length on the flat = 99-975 ft 



LINEAR MEASUREMENT 37 

True length of sub-length on flat 

49-964 
= 1Q0 x 99-9747 = 49-952 

Sag correction for 50 ft (c oc / 3 ) 

= l/8c, = -0-005 
True length between supports = 49- 947 ft 
Alternatively, by Eq.(1.48) 

50 x 0-01 2 
° = 100 x 24 x 10» (5 ° " 10 ° ') 

= -0-018 ft 

."• true length between supports = 49-964 - 0-018 

= 49-946 ft. 

Example 1.23 A copper transmission line, %in. diameter, is stretched 
between two points, 1000ft apart, at the same level, with a tension of 
ftton,when the temperature is 90 °F. It is necessary to define its 
limiting positions when the temperature varies. Making use of the cor- 
rections for sag, temperature and elasticity normally applied to base 
line measurements by tape in catenary, find the tension at a tempera- 
ture of 10 °F and the sag in the two cases. 

Young's modulus for copper 10 x 10 6 lbf/in 2 , its density 555 lb/ft 3 , 
and its coefficient of linear expansion 9-3 x 10~ 6 per °F. 

(L.U.) 
The length of line needed = 1000 + 81 

where 81 = added length due to sag 
= H ' 2/3 

24 r 2 

w = 7Tr z p lbf/ft 

3-142 x 0-25 2 x 555 



144 



0-757 lbf/ft 



s; 0-757 2 x 1000 3 

ol = = 19-037 ft 

24 x 1120 2 i»iw/K 



Total length of wire = 1019-037 ft 

w Z 2 
amount of sag v = 

y &T 

_ 0-759 x 1019 ; 
8 x 1120 



87-73 ft 



38 



SURVEYING PROBLEMS AND SOLUTIONS 



when temperature falls to 10 °F, 
Contraction of wire = Lat 

= 1000 x 9-3 x 10 -6 x (90-10) = 0-758 ft 
new length of wire = 1019-037-0-758 = 1018-279 ft 



as SI oc 
T 2 = T 



T 2 
2 8l ± 



r 2 = 1120 



/19-037 
V18-279 



= 1120 x 1-0205 = 1142 lbf 



Amount of sag at 10 °F 



(y ool) 



T 

= y, x li 



= 87-73 x ii?| = 86-03 ft 



1.56 Reduction to mean sea level (Fig. 1.29) 



If the length at mean sea level 
is L and h = height of line above 
or below mean sea level, then 

L - R 
L 



L = 



R±h 
R±h 




If L = l m T c, then 



Fig. 1.29 Reduction to mean sea level 



l„R 



c ~ /mT /f±T 



(1.53) 



= T 



R±h 



LINEAR MEASUREMENT 



39 



As h is small compared with R, c = ± ~^— 

R 

If R ^ 3960 miles, 



(1.54) 



100ft 



3960 x 5280 

4-8/j xl0~ 6 per 100 ft 



(1.55) 



1.57 Reduction of ground length to grid length 

The local scale factor depends on the properties of the projection. 

Here we will consider only the Modified Transverse Mercator pro- 
jection as adopted by the Ordnance Survey in the British Isles. 
Local scale factor (F) 



■(■•=) 



F = F o\l + „ 



(1.56) 



where F Q = the local scale factor at the central meridian, 
E = the Easting in metres from the true origin, 
p = the radius of curvature to the meridian, 
v = the radius of curvature at right angles to the meridian. 
Assuming p~v = R, then 



F = F 



I 1 + ~w) 



For practical purposes, 

F ~ F (l + 1-23 E 2 x 10" 8 ) 

- 0-9996013(1 + 1-23E 2 x 10~ 8 ) 
N.B. E = Eastings - 400 km. 



(1.57) 

(158) 
(1.59) 




0-9996 



Sub-parallel 



180km 



Central meridian 



L.S.F. 



| ( LS - E -=^) 




0-9996 



Local scale 
error 1-0 



T 



0.9996 



Sub-parallel 



180km Distance from 
CM. in km 



Fig. 1.30 



40 



SURVEYING PROBLEMS AND SOLUTIONS 



The local scale error as shown on the graph approximates to 



2R'< 



Example 1.24 Calculate (a) the local scale factors for each corner of 
the grid square TA (i.e. grid co-ordinates of S.W. corner 54), (b) the 
local scale factor at the centre of the square, (c) the percentage error 
in each case if the mean of the square corners is used instead. 

w fc £ 

& s 8 

0> O) o 

o> o> p o 



TA 
(54) 



£<km) 



8 jp 

Fig. 1.31 

See Chapter 3, page 160. 

(a) (i) At the S.W. corner, co-ordinates are 500 km E, i.e. 100km 

E of central meridian. 
Therefore, from Eq.(1.58), 

L.S.F. = 0-9996013(1 + 100 2 x 1-23 x 10" 8 ) 
= 0-999601 + 0-000123 = 0-999724 

(ii) At S.E. corner, co-ordinates are 600 km E, i.e. 200 km E of 
CM. 

.-. L.S.F. = 0-9996013(1 + 200 2 x 1-23 x 10~ 8 ) 

= 0-999601 + 2 2 x 0-000123 

= 0-999601 + 0-000 492 = 1*000 093 

(b) At centre of square, 150 km from central meridian, 

L.S.F. = 0-999 601 + 1'5 2 x 0-000123 = 0-999878 

(c) (i) % error at S.W. corner 

0-999724 - y 2 (-999724 + 1-000 093) 



0-999 724 



x 100 



= 0-018% 



LINEAR MEASUREMENT 41 



(ii) % error at S.E. corner 

1-000 093 - 0-999908 
1-000093 

(iii) % error at centre 

= 0-999878 - 0-999908 
0-999878 



100 = 0-019% 



x 100 = 0-003% 



Example 1.25 Calculate the local scale factors applicable to a place 
E 415 km and to coal seams there at depths of 500 ft, 1000 ft, 1500 ft 
and 2000 ft respectively. 

Local radius of the earth = 6362-758 km 
L.S.F. = 0-9996013(1 + E 2 x 1-23 x 10" 8 ) (Eq. 1.59) 
Correction of length to mean sea level 

~ R - h 

at 500ft L = / 6362-758 

m 6362-758 - (500 x 0-3048 x 10~ 3 ) 
6362-758 



= /, 



6362-758 - 0-152 



. 6362-758 „ nnnnn , 
L 6362^606 ' 1'MOMH 



at 1000 ft T , 6362-758 



m 6362-758 - 0-304 



at 1500 ft L = I 6362 ' 758 

m 6362-758 - 0-457 

, 6362-758 



at 2000 ft L = / 6362-758 



6362-758 - 0-610 

, 6362-758 
■ '""6362048 " 1 - 000096 '' 



At mean sea level, Easting 415 km, 



42 SURVEYING PROBLEMS AND SOLUTIONS 

L.S.F. = 0-9996013 [1 + (415 - 400) 2 x 1-23 x 10" 8 ] 
= 0-9996040 

at 500 ft below, 

L.S.F. = 0-999604 x 1-000024 = 0-999628 

at 1000 ft below, 

L.S.F. = 0-9996040 x 1-000048 = 0-999652 

at 1500 ft below, 

L.S.F. = 0-9996040 x 1-000072 = 0-999676 

at 2000 ft below, 

L.S.F. = 0-999604 x 1-000096 = 0-999700. 

Example 1.26 An invar reference tape was compared with standard on 
the flat at the National Physical Laboratory at 68 °F and 201bf tension 
and found to be 100-024 Oft in length. 

The first bay of a colliery triangulation base line was measured in 
catenary using the reference tape and then with the invar field tape at 
a temperature of 60° F and with 20 lbf tension. The means of these 
measurements were 99*876 3 ft and 99*912 1 ft respectively. 

The second bay of the base line was measured in catenary using 
the field tape at 56 °F and 20 lbf tension and the resulting mean mea- 
surement was 100*213 5 ft. 
Given: 

(a) the coefficient of expansion for invar = 3-3 x 10" 7 , 

(b) the weight of the tape per foot run = 0-00824 lbf, 

(c) the inclination of the second bay = 3° 15' 00", 

(d) the mean height of the second bay = 820 ft A.O.D. 

Assuming the radius of the earth to be 20 890 000 ft, calculate the 
horizontal length of the second bay reduced to Ordnance Datum. 

(M.Q.B./S) 

To find the standardised length of the field tape . 

Reference tape on the flat at 68 °F = 100-024 Oft. 

Temperature correction 

= 100 x 3-3 x 10" 7 x (60 - 68) 

= -0-000 264 i.e. -0-0003 
.*. Reference tape at 60 °F 

= 100-024 - 0-0003 

= 100-023 7 ft. 



LINEAR MEASUREMENT 



43 



Thus the standardisation correction is + 0*023 7 ft per 100 ft. 

-w 2 / 3 



Sag correction 



24T 2 
-(8-24 x 10~ 3 ) 2 x 100 3 



- 8-24 2 
9600 



24 x 20 2 
= -0*0071 ft 



The length of 100 ft is acceptable in all cases due to the close approxi- 
mation. 

The true length of the first bay thus becomes 

99-8763 - 0-0071 + 0-0237 = 99-892 9 ft. 

The field tape applied under the same conditions when corrected for 
sag gives 

99-9121 - 0-0071 = 99-905 Oft. 
The difference represents the standardisation correction 

99-9050 - 99-8929 = +0-022 lft. 
The corrections may now be applied to the second bay . 
Standardisation 

100-21 



c = + 0-0221 x 



99-91 



+ 
0-0221 



(the proportion is not necessary because of the 
close proximity) 

Temperature, L.a.(t m - t a ) 

c = 100 x 3-3 x 10" 7 x (56 - 60) 
Sag, 

As before as length a 100 ft 
Slope, L (1 - cos 6) 

c = - 100-21(1- cos 3° 15') 
= - 100-21 x 0-00161 
Sea level, - Ih/R 

c = - 100 x 820/20890000 



0-022 1 



0-0001 
0-0071 

0-1613 
0-003 9 



0-1724 
0-022 1 
0-1503 



44 



SURVEYING PROBLEMS AND SOLUTIONS 



Horizontal length reduced to sea level 

= 100-2135 - 0-1503 
= 100-063 2 ft. 



Example 1.27 The details given below refer to the measurement of 
the first '100ft' bay of a base line. Determine the correct length of the 
bay reduced to mean sea level. 

With the tape hanging in catenary at a tension of 20 lbf and at a 
mean temperature of 55° F, the recorded length was 100-082 4 ft. The 
difference in height between the ends was 1-52 ft and the site was 
1600ft above m.s.l. 

The tape had previously been standardised in catenary at a tension 
of 15 lbf and at a temperature of 60 °F, and the distance between zeros 
was 100-042 ft. R = 20890000 ft. Weight of tape/ft = 0-013 lbf. Sec- 
tional area of tape = 0-005 6 in 2 , E = 30 x 10 6 lbf/ in 2 . Temperature 
coefficient of expansion of tape = 0-00000625 per 1°F. 

(I.C.E.) 



Corrections . 

Standardisation 

Tape is 100-042 ft at 15 lbf tension and 60 °F. 
.-. c = 0-042 ft per 100 ft 

Temperature 

c = L . a . (t m - t s ) 

= 100 x 6-25 x 10" 6 x (55-60) = -0-0031 

Tension 

_ L(T m -T 9 ) 



A.E 

100 x (20 - 15) 
0-0056 x 30 x 10 6 



= +0-003 



Slope 



c = 



21 

1-52 2 
200 



0-0116 



Correction 

+ 



0-0420 



0-003 



0-003 1 



0-0116 



Sag 



c = difference between the corrections for 
field and standard tensions 



LINEAR MEASUREMENT 



45 



C = - 



W 2 I 
24 



TJ 



1 



W 



c = 



wl = 0-013 x 100 = 1-3 lbf 



1-32 2 x 100 



24 
= +0-013 7 



15 2 - 20 2 
15 2 x 20 2 



Height 



c = 



hl_ 
R 

1600 x 100 
20890000 



= -0-007 7 



0-013 7 



+ 0-0587 
-0-0224 



+ 0-0363 



0-0077 



0-022 4 



Measured length + 100-0824 
Total correction + 0-0363 
Corrected length 100-118 7 ft 

1.6 The Effect of Errors in Linear Measurement 

If the corrections previously discussed (pp. 23-40) are not applied 
correctly, then obviously errors will occur. Any errors within the for- 
mulae produce the following effects. 

1.61 Standardisation 

Where a tape is found to deviate from standard, the error SI can be 
corrected in the normal way or by altering the standard temperature as 
previously suggested. 

1.62 Malalignment and deformation of the tape (Figs. 1.32 and 1.33) 

(a) Malalignment. If the end of the tape is out of line by an amount d 
in a length /, the error will be 



e.g., if d= 3 in. and /= 100ft, 



21 



(1.60) 



46 SURVEYING PROBLEMS AND SOLUTIONS 



- = ^L = 0-000 3 ft, JL 

200 



i.e. 1 in 330000. 

e 

Fig. 1.32 Malalignment of the tape 

(b) Deformation in the horizontal plane. If the tape is not pulled 
straight and the centre of the tape is out of line by d, then 

d d 2d ,„ _„ N 1/_| ,. 

e z = TTv"" -77 v = T" < L61) Jtl "Sp &L 



4X1) ' 



&*=. 1 =**o 

Fig. 1.33 Deformation of the tape 

e.g., if d = 3 in. and 1= 100ft, 

e 2 = 4 x e, = 0-00123, i.e. 1 in 80000. 

(c) Deformation in the vertical plane. This is the same as (b) but more 

difficult to detect. Any obvious change in gradient can be allowed for 

by grading the tape or by measuring in smaller bays between these 

points. 

N.B. In (a) and (b) alignment by eye is acceptable for all purposes 

except very precise work. 

1.63 Reading or marking the tape 

Tapes graduated to 0-01 ft can be read by estimation to give a pro- 
bable error of ± 0-001 ft. 

Thus if both ends o f the tape are read simultaneously the probable 
error in length will be V(2 x 0001 2 ), i.e. 0-001 x y/2, i.e. ±0-001 4 ft. 

Professor Briggs suggests that the error in setting or marking of 
the end of the tape is 3 times that of estimating the reading, i.e. 
±0*003 ft per observation. 

1.64 Errors due to wrongly recorded temperature 
From the correction formula c = I. a. (t m - t^), 

dc = ladtn (1.62) 

and j. = aStn (1.63) 

It has been suggested from practical observation that errors in re- 
cording the actual temperature of the tape for ground and catenary mea- 
surement are ± 5°F and ± 3°F respectively. 



LINEAR MEASUREMENT 47 

If the error is not to exceed 1/10000, then from Eq. 1.63 
8c 1 



•= a 8t, 

1UUUU 

i.e. 8t = 



I 10000 

1 



10000 a 
If a = 6-5 x 10" 6 per deg F, then 

10 6 
8t m = — ~ 15.4 

6'5 X 10* 

Thus 5° produces an error of ~ 1/30000, 
3° produces an error of ^ 1/50000. 

1.65 Errors due to variation from the recorded value of tension 

These may arise from two sources: 

(a) Lack of standardisation of tensioning apparatus. 

(b) Variation in the applied tension during application (this is sig- 
nificant in ground taping). 

From the correction formula (1.39) c = H Tm ~ T *) _ LT 

AE ~~AE 



differentiation gives g c = L8T, 

AE 



(1.64) 
(1.65) 



i.e. 



c " r 

8c 8T m 

T = IE < 166 > 



If the error is not to exceed say 1 in 10 000, then 

L AE 10000 

• <s T AE 

i-e- oi m - 1O(M)0 

If A = 0-003 in 2 , E = 30 x 10 a lbf/in 2 , then 
- T 0-003 x 30 x 10 6 

8 m = — w — '~ = 91bf - 

i.e., an error of 1/10000 is produced by a variation of 91bf, 
an error of 1/30 000 is produced by a variation of 3 Ibf . 
The tape cross-section is x / 2 in. wide to give A = 0*003 in 2 . If the 
width of the tape be reduced to Kin. then, if the other dimensions 



48 SURVEYING PROBLEMS AND SOLUTIONS 

remain constant, the cross-sectional area is reduced to %A = 0*000 8 in 2 . 

In this case a variation of 3 lbf will produce an error of 1/10 000 
and the accuracy will be reduced as the cross-sectional area dimini- 
shes. 

1.66 Errors from sag 

Where the tape has been standardised on the flat and is then used 
in catenary with the measuring heads at different levels, the approxi- 
mation formula is given as 

-/ 3 w 2 cos 2 6 

c = (1.50) 

24I 2 

where 6 is the angle of inclination of the chord between measuring 
heads. The value of cos 2 becomes negligible when 6 is small. 
The sources of error are derived from: 

(a) an error in the weight of the tape per unit length, w, 

(b) an error in the angular value, B, 

(c) an error in the tension applied, T. 
By successive differentiation, 

dc w = - 2/ 3 wcos 2 flg W (1 67) 

24 T 2 

2cdW (1.68) 



i.e. 



w 
8c„, 28w 



c w 



(1.69) 



This may be due to an error in the measurement of the weight of the 
tape or due to foreign matter on the tape, e.g. dirt. 

-l 3 w 2 
8c e = — 2 sin20S0 



See 
c 



Sc T = 



= 2c tan 6 86 
= 2tan0S0 

2/ 3 w 2 cos 2 0ST 



24T : 
2c8T 



T 

8c T _ -28T 

c " T 



(1.70) 


(1.71) 


(1.72) 


(1.73) 


(1.74) 


(1.75) 



LINEAR MEASUREMENT 49 

The compounded effect of a variation in tension gives 

2l*w 2 cos 2 (9 8T 18T 

24T 2 + ~AE < L76 ) 

Example 1.28 If / = 100 ft, w = 0-01 ± 0-001 lbf per ft, $ = 2° ± 10" 
T = 10 ± 1 lbf, 



100 3 x 0-01 2 x cos 2 2° cos 2 2° 

~24 



C = 24xl0 2 = ^lf~ = 0>04161ft - 



Then~ 2c 8w 

w = ~w~ = 2 x °* 04161 x O'l = 0-008 32 ft 

i.e. 10% error in weight produces an error of 1/12000. 

« 2c tan fl 86" 0-08322 

806 = 206 265 = "2^6167 x °-° 524 x 10 

= 0-000 000 21 ft. 
This is obviously negligible. 

Sc t = ~y- = 0-083 22 xO-U 0-008 322 ft 
i.e. 10% error in tension produces an error of 1/12000. 

Example 1.29 A base line is measured and subsequent calculations 
show that its total length is 4638-00 ft. It is later discovered that the 
tension was recorded incorrectly, the proper figure being 10 lbf less 
than that stated in the field book, extracts from which are given below. 
Assuming that the base line was measured in 46 bays of nominal 
length 100 ft and one bay of nominal length 38 ft, calculate the error 
incurred in ft. 

Extract from field notes 

Standardisation temperature = 50 °F 

Standardisation tension = 20 lbf 

Measured temperature = 45 op 

Measured tension _ 40 lbf 

Young's modulus of tape = 30 x 10 6 lbf/in 2 

Cross-sectional area of tape = 0-125 in. x 0-05 in.- 

Weight of lin 3 of steel = 0-28 lbf. 

(N.U.) 
Weight of steel tape per ft = 0-125 x 0-05 x 12 x 0-28 = 0-021 lbf. 



50 SURVEYING PROBLEMS AND SOLUTIONS 

From Eq. (1.39) c = L(Jm ' J s) 

Then the error due to wrongly applied tension = c - c' 

= L(T m - r.) _ L(T^ - T s ) 
AE AE 

= M. (T - T') 
AE K m m) 

where T m = true applied tension, 

T' m = assumed applied tension. 



Error 



4638 ( 30 - 40 > . . -0-24736ft. 
0-125 x 0-05 x 30 x 10 6 



-W z i 
From Eq. (1.46) correction for sag c = — — -j 

.'. Error due to wrongly applied tension 
= c - c, 

w!i/i _ i_\ 

24 \T Z Tfj 
W = 100 x 0-021 = 2-1 lbf 
Error for 100 ft bay 

2-1 2 x 100 /Jl_ _ _1_\ 
24 Uo 2 40 2 / 

441 / l600 - 900 \ 
= ~ 24 \l600 x 900/ 

_ 441 / 700 \ 
24 \1 440 000/ 

= -0-00893 
Error for 46 bays = -0-410 78 ft 

Error for 38 ft bay 

W = 38 x 0-021 = 0-798 lbf 

_ _ 0-798 2 x 38 / 700 \ 
n0t ~ 24 \l 440 000/ 

= - 0-000 49 ft 

.-. Total error for sag = -0-411 27 ft 

Total error for tension = - 0-247 36 ft 



LINEAR MEASUREMENT 51 

Total error = - 0-658 63 ft 
i.e. Apparent reduced length is 0*658 6 ft too large. 

1.67 Inaccurate reduction to the horizontal 

The inclined length may be reduced by obtaining 

(a) the difference in level of the measuring heads or 

(b) the angle of inclination of the tape, 
(a) The approximation formula is given as 



(Eq. 1.15) 


d 2 
C =2l 




...) 


Adopting the first term only, 


from the differentiation 




8c 


_ d8d 
I 

2c 8d 
d 






8c 
c 


28d 
d 





(1.77) 
(1.78) 

(1.79) 

If 8d/d= 1%, when / = 100 and d= 5 ± 0-05 ft, 

* 2 x 5 2 x 0-01 

8c = o T7^ = ± 0-002 5 f t i.e. 1 in 40000. 

2 x 100 

As the difference in level can be obtained without difficulty to 
± 0-01 ft, 

8c = ± 0-000 5 ft, i .e . 1/200 000 . 
(b) By trigonometrical observations (Eq. 1.12), 

c = 1(1 - cos0) 
Then 8c = /sin0S0 (1.80) 

_ /sinflgfl" 

" 206265 C1,81) 

i.e. 8c oc sin0 86 (1.82) 

If / = 100, B = 30° ± 20", 

8c = ± 10 ° * °; 5 * 2 ° - ±0-0048ft i.e. 1/20000. 
2Uo 265 

The accuracy obviously improves as 6 is reduced. 

As the angle of inclination increases the accuracy in the measure- 
ment of 6 must improve. 



52 SURVEYING PROBLEMS AND SOLUTIONS 

1.68 Errors in reduction from height above or below mean sea level 

From the formula 

Ih 

by differentiating 8c = (1.83) 

R. 

cSh 
= T£- (1.84) 

The % error in the correction is equal to the % error in the height above 
or below M.S.L. 

1.69 Errors due to the difference between ground and grid distances 

Local scale factor is given by Eq.(1.59) 

0-9996013(1 + 1-23 E 2 x 10~ 8 ) 

where E is the distance in km from the central meridian (i.e. the East- 
ings - 400 km). 

As this amounts to a maximum of 0*04% it is only effective in pre- 
cise surveys . 

Exercises 1(b) 

10. A 300ft tape has been standardised at 80 °F and its true length 
at this temperature is 300*023 ft. A line is measured at 75 °F and re- 
corded as 3486*940 ft. Find its true length assuming the coefficient of 
linear expansion is 6*2 x 10 -6 per deg F. 

(Ans. 3487- 10 ft) 

11. A base line is found to be 10560 ft long when measured in caten- 
ary using a tape 300ft long which is standard without tension at 60 °F. 
The tape in cross-section is 1/8 x 1/20 in. 

If one half of the line is measured at 70 °F and the other half at 
80 °F with an applied tension of 501bf, and the bays are approximately 
equal, find the total correction to be applied to the measured length. 

Coefficient of linear expansion = 6*5 x 10 -6 per deg F. 

Weight of 1 in 3 of steel = 0-28 lbf . 

Young's modulus = 29 x 10 6 lbf/in 2 . 

(Ans. -3 -042 ft) 

12. A 100 ft steel tape without tension is of standard length when 
placed on the ground horizontally at a temperature of 60 °F. The cross- 
sectional area is 0*0103 in 2 and its weight 3*49 lbf, with a coefficient 
of linear expansion of 6*5 x 10~ 6 per deg F. 

The tape is used in the field in catenary with a middle support such 



LINEAR MEASUREMENT 53 

that all the supports are at the same level. 

Calculate the actual length between the measuring heads if the 
temperature is 75 °F and the tension is 20 lbf. (Assume Young's mod- 
ulus 30 x 10 6 lbf/in 2 ). 

(Ans. 99-985 lft) 

13. A nominal distance of 100 ft was set out with a 100 ft steel tape 
from a mark on the top of one peg to a mark on the top of another, the 
tape being in catenary under a pull of 20 lbf and at a mean temperature 
of 70 °F. The top of one peg was 0*56 ft below the top of the other. 
The tape had been standardised in catenary under a pull of 25 lbf at a 
temperature of 62 °F. 

Calculate the exact horizontal distance between the marks on the 
two pegs and reduce it to mean sea level. The top of the higher peg 
was 800ft above mean sea level. 

(Radius of earth = 20:9 x 10 6 ft ; density of tape 0*28 lbf/in 3 ; 
section of tape = 0*125 x 0*05 in.; Young's modulus 30 x 10 6 lbf/in 2 ; 
coefficient of expansion 6' 25 x 10" 6 per 1° F) 

(I.C.E. Ans. 99-980 4 ft) 

14. A steel tape is found to be 299-956 ft long at 58 °F under a ten- 
sion of 12 lbf. The tape has the following specifications: 

Width 0-4 in. 

Thickness 0-018 in. 

Young's modulus of elasticity 30 x 10 5 lbf/in 2 
Coefficient of thermal expansion 6-25 x 10" 6 per deg F. 
Determine the tension to be applied to the tape to give a length of 
precisely 300 ft at a temperature of 68 °F. 

(N.U. Ans. 30 lbf) 

15. (a) Calculate to three decimal places the sag correction for a 
300 ft tape used in catenary in three equal spans if the tape weighs 
1 lb/100 ft and it is used under a tension of 20 lbf. 

(b) It is desired to find the weight of a tape by measuring its sag 
when suspended in catenary with both ends level. If the tape is 100 ft 
long and the sag amounts to 9-375 in. at mid-span under a tension of 
20 lbf, what is its weight in ozf per 100 ft? 

(N.U. Ans. 0-031 ft, 20 ozf) 

16. Describe the methods used for the measurement of the depth of 
vertical mine shafts and discuss the possible application of electronic 
distance measuring equipment. 

Calculate the elongation of a shaft measuring tape due to its own 
weight at (1) 1000 ft and (2) 3000 ft, given that the modulus of elastici- 
ty is 30 x 10 6 lbf/in 2 } weight of the tape 0-051bf/ft run, and the cross- 
sectional area of the tape 0-015 in 2 . 

(N.U. Ans. 0-055 6 ft, 0-500 ft) 



54 SURVEYING PROBLEMS AND SOLUTIONS 

17. (a) Describe the measuring and straining tripods used in geodelic 
base measurement. 

(b) The difference between the readings on a steel tape at the 
terminals of a bay between which it is freely suspended was 94*007 ft, 
the tension applied being 201bf, the temperature 39*5 °F, and the height 
difference between the terminals 5*87 ft. The bay was 630 ft above mean 
sea level. 

If the tape, standardised on the flat, measured correctly at 68 °F 
under lOlbf tension, and its weight was 0*0175 lbf per ft, its coefficient 
of expansion 0*62 x 10~ 8 per deg F and its coefficient of extension 
0*67 x 10" 5 per lb, calculate the length of the bay reduced to mean sea 
level. (Radius of earth = 20-9 x 10 6 ft). 

(L.U. Ans. 93-784 ft) 

18. The steel band of nominal length 100ft used in the catenary mea- 
surement of a colliery base line, has the following specification: 

(i) Length 100*025 ft at 10 lbf tension and 68 °F. 
(ii) Sectional area 0*004 in 2 
(iii) Weight 22 ozf . 

(iv) Coefficient of linear expansion 6*25 x 10~ 6 per deg F. 
(v) Modulus of elasticity 30 x 10* lbf/in 2 . 

The base line was measured in 10 bays and the undernoted obser- 
vations were recorded in respect of the first five which were of average 
height 625 ft above Ordnance Datum. 





Observed 


Bay 


Bay Level 


Tension 


Bay 


Length 


Temperature 


Difference 


Applied 


1 


100*005 


52 °F 


0*64 


20 lbf 


2 


99*983 


54°F 


1*23 


20 lbf 


3 


100*067 


54 °F 


0*01 


20 lbf 


4 


100*018 


58 °F 


0*79 


20 lbf 


5 


99*992 


60°F 


2*14 


20 lbf 



Correct the bays for standard, temperature, tension, sag, slope, 

and height above Ordnance Datum and compute the corrected length of 

this part of the base line. Take the mean radius of the earth to be 

20890000 ft. 

(M.Q.B./S Ans. 500'044ft) 

19. The steel band used in the catenary measurement of the base line 
of a colliery triangulation survey has the undernoted specification: 
(i) length 50*000 3 m at a tension of 25 lbf at 60° F. 

(ii) weight 2*5 lbf 

(iii) coefficient of linear expansion 6*25 x 10~ 8 per deg F. 



LINEAR MEASUREMENT 55 

The undernoted data apply to the measurement of one bay of the 
base line: 

(i) length 50-002 7 m 
(ii) mean temperature 53 °F 
(iii) tension applied 25 lbf 

(iv) difference in level between ends of bay 0*834 m 
(v) mean height of bay above mean sea level 255*4 m. 
Correct the measured length of the bay for standard, temperature, 
sag, slope, and height above mean sea level. Assume the mean radius 
of the earth is 6-37 x 10 6 m. 

(M.Q.B./S Ans. 49-971 m) 

20. A base line was measured with an invar tape 100 ft long which 
had been standardised on the flat under a tensile load of 15 lbf and at 
a temperature of 60 °F. Prior to the measurement of the base line the 
tape was tested under these conditions and found to record 0*015 ft too 
much on the standard length of 100 ft. The base line was then divided 
into bays and the results obtained from the measurement of the bays 
with the tape suspended are shown below: 



Bay 


Length 
(ft) 


Difference in level 
between supports (ft) 


Air temperature 
(°F) 


1 


99*768 


2*15 




49*6 


2 


99*912 


1*62 




49*6 


3 


100*018 


3*90 




49*8 


4 


100*260 


4*28 




50*2 


5 


65*715 


0*90 




50*3 



Modulus of elasticity (E) for invar = 22 x 10 6 lbf/in 2 . 

Coefficient of linear expansion of invar = 5*2 x 10~ 7 per deg F. 

Field pull = 25 lbf. 

Cross-sectional area of tape = 0*004 in 2 . 

Weight per ft run of tape = 0*010 2 lbf. 

Average reduced level of base line site = 754*5 ft. 

Radius of earth = 20*8 x 10 6 ft. 

Correct the above readings and determine to the nearest 0*001 ft 
the length of the base line at mean sea level. 

(I.C.E. Ans. 465*397 ft) 

21. The following readings were taken in measuring a base line with 
a steel tape suspended in catenary in five spans: 



56 SURVEYING PROBLEMS AND SOLUTIONS 



pan 


Mean reading 


Difference 


in 


level 


Tension 


Mean 




of tape (ft) 


between Index Marks 


(ft) 


(Ibf) 


temperature (°F) 


1 


100.155 


3-1 








25 


73 


2 


100*140 


0*9 








50 


76 


3 


100«060 


1-2 








25 


78 


4 


100» 108 


3-1 








25 


80 


5 


100-182 


2-0 








25 


80 



The tape reading was 100*005 ft when calibrated in catenary under 
a tension of 25 lbf at a temperature of 65 °F between two points at the 
same level precisely 100 ft apart. 
Other tape constants are: 

width of tape = 0*250 in; thickness of tape = 0*010 in; 
weight of steel = 0*283 lbf/in 3 ; E for steel = 30 x 10 6 lbf/in 2 ; 
coefficient of expansion of steel = 6*2 x 10" per deg F. 

Compute the length of the base line. 

(I.C.E. Ans. 500*568 ft) 

22. A short base line is measured in four bays with a 100ft invar 
band in catenary under a pull of 20 lbf with the following field readings: 

Bay 1 2 3 4 



t°F 


65*2 


64*0 


65*5 


63*8 


hit 


5*08 


1*31 


2*31 


2*13 


I ft 


99*6480 


99*751 7 


99*541 7 


99*9377 



where t is the field temperature, h is the difference in level between 
the ends of each bay and / is the mean reading of the invar band . 

When standardised in catenary under a pull of 20 lbf at 68*5°F the 
standard length of the invar band was 99*999 ft and the mean altitude 
of the base is 221 ft above sea level. If the coefficient of expansion of 
invar is 0*000 0003 per deg F and the radius of the earth is 20*9 x 10* ft 
what is the length of the base line reduced to sea level? 

(I.C.E. Ans. 398*6829 ft) 

Bibliography 

CLARK, D., Plane and Geodetic Surveying, Vols. I and II (Constable) 
HOLLAND, J.L., WARDELL, K. and WEBSTER, A.G., Surveying, Vols. I 

and II. Coal Mining Series (Virtue) 
SMIRNOFF, M.V., Measurement for Engineering and Other Surveys 

(Prentice-Hall) 
BANNISTER, A. and RAYMOND, S., Surveying (Pitman) 
THOMAS, W.N., Surveying, 5th ed. (Edward Arnold) 
BRIGGS, N., The Effects of Errors in Surveying (Griffin) 
MINISTRY OF TECHNOLOGY, Changing to the Metric System (H.M.S.O.) 
ORDNANCE SURVEY publication, Constants, Formulae and Methods 

used in Transverse Mercator (H.M.S.O.) 



2 SURVEYING TRIGONOMETRY 



'Who conquers the triangle half conquers his subject' 

M.H. Haddock 



Of all the branches of mathematics, trigonometry is the most im- 
portant to the surveyor, forming the essential basis of all calculations 
and computation processes. It is therefore essential that a thorough 
working knowledge is acquired and this chapter is an attempt to sum- 
marize the basic requirements. 

2.1 Angular Measurement 

There are two ways of dividing the circle: 

(a) the degree system, 

(b) the continental 'grade' system. 

The latter divides the circle into 4 quadrants of 100 grades each 
and thereafter subdivides on a decimal system. It has little to com- 
mend it apart from its decimalisation which could be applied equally 
to the degree system. It has found little favour and will not be con- 
sidered here. 

2.11 The degree system 





180 



Clockwise rotation used by 
surveyors 



Anticlockwise rotation used by 
mathematicians 



Fig. 2.1 Comparison of notations 
57 



58 



SURVEYING PROBLEMS AND SOLUTIONS 



The circle is divided into 360 equal parts or degrees, each degree 
into 60 minutes, and each minute into 60 seconds. The following sym- 
bols are used:— 

degrees (°) minutes ( ') seconds (") 

so that 47 degrees 26 minutes 6 seconds is written as 

47° 26' 06" 

N.B. The use of 06" is preferred in surveying so as to remove any 
doubts in recorded or computed values. 

In mathematics the angle is assumed to rotate anti-clockwise 
whilst in surveying the direction of rotation is assumed clockwise. 

This variance in no way alters the subsequent calculations but is 
merely a different notation. 

2.12 Trigonometrical ratios (Fig. 2.2) 
Assume radius = 1 

Cotangent© 




a) C 

Versine 
Fig. 2.2 
Sine (abbreviated sin) angle = 4| = 4p = AB = GO 

Cosine (abbreviated cos) angle = 9J* = 93. = OB = GA 

OA 1 

Tangent (abbreviated tan) angle = M. = sin_0 = DC = DC 

OB cos OC 1 



Cotangent 6 (cot 0) = 



_ cos 6_ OB _ FE 



FE 



tan 6 sin 6 AB 



FO 



SURVEYING TRIGONOMETRY 






Cosecant (cosec 0) = —. — ■* = -t-~ = 

sin0 Ad 


OE 
OF 


OE 
1 


Secant (sec 0) - * - °A - 

cos OB 


OD 
OC 


OD 
1 



59 



Versine (vers 0) = 1 - cos = OC - OB 
Coversine (covers 0) = 1 - sin0 = OF - OG. 

N.B. In mathematical shorthand sin -1 x means the angle (x) whose 
sine is ... . 

If OA = radius = 1 

then, by Pythagoras, sin 2 + cos 2 = 1. (2.1) 

sin 2 = 1 - cos 2 (2.2) 

cos 2 = 1 - sin 2 (2.3) 

Dividing Eq. 2.1 by cos 2 0, 

sin 2 cos 2 1 

+ 



cos 2 cos 2 cos 2 

i.e. tan 2 + 1 = sec 2 

tan 2 = sec 2 - 1. (2.4) 

Dividing Eq. 2.1 by sin 2 0, 

sin 2 cos 2 1 



sin 2 sin 2 sin 2 

i.e. 1 + cot 2 = cosec 2 (2.5) 

sin0 = V(l- c os 2 0). (2.6) 

cos0 = V(l-sin 2 0). (2.7) 



(2.8) 
(2.9) 

(2.10) 
(2.11) 



tan0 = 


sin0 sin0 




cos0 V(l-sin 2 0) 




V(l - cos 2 0) 
or = - 5Li - 

COS0 


cos — 


1 1 




sec0 VCi + tan 2 ^) 


sin0 — 


1 1 



cosec \/(l + cot 2 0) 



which shows that, by manipulating the equations, any function can be 
expressed in terms of any other function. 



60 



SURVEYING PROBLEMS AND SOLUTIONS 



2.13 Complementary angles 

The complement of an acute angle is the difference between the 
angle and 90°, 



i.e. 



if angle A 



30 ( 



its complement = 90° - 30° = 60° 



The sine of an angle 
The cosine of an angle 
The tangent of an angle 
The secant of an angle 
The cosecant of an angle 

2.14 Supplementary angles 



cosine of its complement 
sine of its complement 
cotangent of its complement 
cosecant of its complement 
secant of its complement 



The supplement of an angle is the difference between the angle 
and 180°, 

i.e. if angle A = 30° 

its supplement = 180° - 30°= 150°. 

The sine of an angle = sine of its supplement 

cosine of an angle = cosine of its supplement (but a nega- 
tive value) 
tangent of an angle = tangent of its supplement (but a nega- 
tive value) 

These relationships are best illustrated by graphs. 
Sine Graph (Fig. 2.3) 



180 




Fig. 2. 3 The sine graph 
Let the line OA of length 1 rotate anticlockwise. Then the height 
above the horizontal axis represents the sine of the angle of rotation. 

At 90° it reaches a maximum = 1 
At 180° it returns to the axis. 
At 270° it reaches a minimum = -1 
It can be seen from the graph that 



SURVEYING TRIGONOMETRY 



61 



sin 30° = sin (180 - 30), i.e. sin 150° 

= -sin (180 + 30), i.e. -sin 210° 

= -sin (360 -30), i.e. -sin 330° 

Thus the sine of all angles - 180° are +ve (positive) 

the sine of all angles 180 - 360 are -ve (negative). 

Cosine Graph (Fig. 2.4). This is the same as the sine graph but dis- 
placed by 90°. 




Fig. 2. 4 The cosine graph 
cos 30° = -cos (180 -30) = -cos 150° 
= -cos (180 + 30) = -cos 210° 
= +cos(360-30) = + cos 330° 
Thus the cosine of all angles - 90° and 270° -360° are + ve 

90° -270° ate-ve 

Tangent Graph (Fig. 2.5) This is discontinuous as shown. 

■too +oo -hoo 




-oo -oo 

Fig. 2.5 The tangent graph 



62 



SURVEYING PROBLEMS AND SOLUTIONS 



tan 30° = tan (180 +30), i.e. tan 210° 
= -tan (180 -30), i.e. -tan 150° 
= -tan (360 -30), i.e. -tan 330° 
Thus the tangents of all angles 0-90° and 180°- 270° ate + ve 

90°- 180° and 270°- 360° are -ve 
Comparing these values based on the clockwise notation the sign of 
the function can be seen from Fig. 2.6. 




i*£ 1st quodrtmt 0*-90* 




.+ 2nd quadrant 90-180 




• f%ir\* 



3rd quadrant 160-270 




• 4*A*«* 



* 4th quadrant 270-360 



Fig. 2.6 



Let the rotating arm be +ve 6° 
1st quadrant sin 0, 



SURVEYING TRIGONOMETRY 



sin 

COS0, 

tan0, 



—, i.e. + 



±, i 

+ 



2nd quadrant 
2 = (180-0) 



3rd quadrant 
3 = (0-180) 



4th quadrant 
4 = (360-0) 



sin0 2 *» ± 

+ 

cos 2 = jr. 

+ 

tan 2 = — 

sin 3 s= jz 
+ 

COS0 3 ;~ 

tan 3 = 

sin 4 = 
cos0 4 = 
tan 4 = 



.e. + 
.e. + 



i.e. + 

i.e. - 

i*e. - 

i.e. ■-* 

i.e. - 

i.e. + 

i.e. - 



— , i.e. + 



i.e. 



63 



2.15 Basis of tables of trigonometrical functions 

Trigonometrical tables may be prepared, based on the following 
series : 

~ /}3 /J5 /j7 

" H R R (2.12) 



sin0 = 0-£! + i!-£ + 
3! 5! 7! 



3! 5! 7! 

where is expressed as radians, see p. 72. 
and 3! is factorial 3, i.e. 3 x 2 x 1 

5! is factorial 5, i.e. 5x4x3x2x1. 

cos0 = 1 _ jf? + §1 _ §1 + _.. (2.13) 

2! 4! 6! 

This information is readily available in many varied forms and 



64 



SURVEYING PROBLEMS AND SOLUTIONS 



to the number of places of decimals required for the particular prob- 
lem in hand. 

The following number of places of decimals are recommended:— 

for degrees only, 4 places of decimals, 

for degrees and minutes, 5 places of decimals, 

for degrees, minutes and seconds, 6 places of decimals. 



2.16 Trigonometric ratios of common angles 

The following basic angles may be calculated. 

B 





' 30° 


30° \ 


2/ 




\ 2 

>/3 \ 


/\ 60 ° 




^ 6oy \ 



A 1 D 1 C 

Fig. 2.7 Trigonometrical ratios of 30° and 60 

From the figure, with BD perpendicular to AC, 

Let AB = BC = AC = 2 units, 

then AD = DC = 1 unit, 

by Pythagoras BD = y/(2 2 -l z ) = y/3. 

Thus 



sin 30° =1=0-5 = cos 60° 
2 



sin 60° = 
tan 30° = 



V3 



1-732 



0-8660 



cos 30° 



T ~ 2 

_L = V? 1-732 

V3 3 3 

= 0-577 3 = cot 60° 



tan 60° = ^ 



= 1-7320 = cot 30° 



Similarly values for 45° may be obtained. 

Using a right-angled isosceles triangle where AC = BC = 1, 

by Pythagoras AB = yj{\ 2 + l 2 ) = V 2 



SURVEYING TRIGONOMETRY 



65 





Fig. 2.8 Trigonometrical ratios of 45° 
1 = )il 1-414 2 



0-7071 = cos 45° 



cot 45° 



It can now be seen from the above that 

sin 120° = sin (180 -120) = sin 60° = 0-8660 

/hereas cos 120° = -cos (180 - 120) = -cos 60° = -0-5 

tanl20° = -tan (180 - 120) = -tan60° = -1*7320 

vlso sin 210° = -sin (210 -180) = -sin 30° = -0-5 

cos240° = - cos (240 - 180) = -cos60° = -0-5 

tan 225° = +tan (225 - 180) = tan 45° = 1-0 

sin330° = - sin (360 - 330) = -sin30° = -0-5 

cos 315° = +cos(360-315) = + cos 45° = 0-7071 

tan 300° = - tan (360 - 300) = -tan 60° = -1-7320 

17 Points of the compass (Fig. 2.9) 

These are not used in Surveying but are replaced by Quadrant (or 
uadrantal) Bearings where the prefix is always N or S with the suffix 
or W, Fig. 2.10. 

g. NNE = N 22° 30' E. 

ENE = N 67° 30' E. 
ESE = S 67° 30' E. 
SSE = S 22° 30' E. 
SW = S 45° 00' W. 
NW = N 45° 00' W. 



66 



SURVEYING PROBLEMS AND SOLUTIONS 




Fig. 2.9 Points of the compass 



NW»N45°W 



SW = S45° W 



NNE 

= N22° 30' E 




ENE 

= N67° 30'E 



ESE 
=S67° 30'E 



SSE 

=S22° 30'E 



SURVEYING TRIGONOMETRY 



67 



2.18 Easy problems based on the solution of the right-angled triangle 

N.B. An angle of elevation is an angle measured in the vertical plane 
where the object is above eye level, i.e. a positive vertical angle, 
Fig. 2.11. 

An angle of depression is an angle measured in the vertical plane 
where the object is below eye level, i.e. a negative vertical angle. 




Fig. 2. 11 Vertical angles 

In any triangle there are six parts, 3 sides and 3 angles 

The usual notation is to let the side opposite the angle A be a 

etc, as shown in Fig. 2.12. 

The following facts are thus 

known about the given right-angled 

triangle ABC. 

Angle C = 90° 

Angle A + Angle fi = 90° 

c 2 = a 2 + b z (by Pythagoras) 

sin A = SL 
c 

cos A = k 
c 



tan A = £ 
b 




Fig. 2.12 



68 



SURVEYING PROBLEMS AND SOLUTIONS 



tan A = 



a c 
- x -r 
c b 



i.e. sin A ~- cos A. 

To find the remaining parts of the triangle it is necessary to know 
3 parts (in the case of the right-angled triangle, one angle = 90° and 
therefore only 2 other facts are required). 

Example 2.1. In a right-angled triangle ABC, the hypotenuse AB is 
10 metres long, whilst angle A is 70°. Calculate the remaining parts 
of the triangle. 

As the hypotenuse is AB (c) 
the right angle is at C (Fig. 2.13). 

then 



a 
c 


= 


sin 70° 


a 


= 


c sin 70° 




= 


10 sin 70° 




= 


10 x 0-939 69 




= 


9-397 metres 


b 
c 


= 


cos 70° 


b 


= 


c cos 70° 




= 


10 x 0-34202 




= 


3-420 metres 




Check 



Fig. 2. 13 



AngleS = 90° - 70° = 20° 

A = tanS = l'*®l = 0-363 97 
a 9-396 9 



i.e. angle B = 20° 00' 



Example 2.2 It is necessary to climb a vertical wall 45 ft (13-7 m) 
high with a ladder 50 ft (15-2 m) long, Fig. 2.14. Find 

(a) How far from the foot of the wall the ladder must be placed, 

(b) the inclination of the ladder 

|| = sin^ = 0-9 
50 

angle A = 64° 09' 30" 

thus angle B = 25° 50' 30" 



£ = cos A 



thus 



b = 50 cos 64° 09' 30" = 21-79 ft (6-63 m) 
Ans. (a) 21-79 ft (b) 64° 09' 30" from the horizontal. 



SURVEYING TRIGONOMETRY 
B 



45' 



69 




Fig. 2.14 

Example 2.3. A ship sails 30 miles (48-28 km) on a bearing N 30° E. 
It then changes course and sails a further 50 miles (80'4 km) N 45° W. 

Find (a) the bearing back to its starting point, 
(b) the distance back to its starting point. 
N.B. See chapter 3 on bearings 

To solve this problem two triangles, ADB and BCE, are joined 
to form a resultant third ACF (Fig. 2.15). 

In triangle ADB, AB is N30°E 
30 miles (48-28 km). The distance 
travelled N = AD. 



but 



4§ = cos 30° 
AB 



then AD = 30 cos 30° 

= 25-98 miles (41 -812 km) 
The distance travelled E = DB 

but M = sin 30° 
AB 

.-. DB = 30 sin 30° 

= 15-00 miles (24 -140 km) 

Similarly in triangle BCE the dis- 
tance travelled N = BE 

but BE = BC cos 45° 

= 50 cos 45° 

= 35-35 miles (56-890 km) 

The distance travelled W (CE) = 
The distance travelled N = 35-35 
miles as the bearing = 45° 
(sin 45° = cos 45°). 
In resultant triangle ACF, 




70 



SURVEYING PROBLEMS AND SOLUTIONS 



CF = CE - DB = 35-35 - 15-00 = 20-35 miles (32-750) 
AF = AD + BE = 25-98 + 35-35 = 61-33 miles (98-702) 

tantf = CF = JP-^I = 0-33181 
AF 61-33 

6 = 18° 21 ' 20" .-. bearing A C = N 18° 21' 20" W. 



AC 



AF 
cos 6 



61-33 



cos 18° 21' 20 



'-, — Tl = 64-62 miles (104-0 km) 



Example 2.4. An angle of elevation of 45° was observed to the top 
of a tower. 42 metres nearer to the tower a further angle of elevation 
of 60° was observed. 

Find (a) the height of the tower, 

(b) the distance the observer is from the foot of the tower. 




Fig. 2. 16 



InF] 


ig.2.16, 

AC 

H 


= 


cot A 


i.e. 


AC 


= 


H cot A 


also 


BC 
H 


= 


cotS 


i.e. 


BC 


= 


H cot B. 




AC - BC 




AB = ft (cot ,4 - cotB) 
H AB 




cot A - cot B 



SURVEYING TRIGONOMETRY 71 

42 

cot 45° -cot 60° 

42 
1-0-577 4 



— 


0-4226 - 99 " 38ln 


BC = 


H cot B 


= 


99-38 cot 60° 


= 


99-38 x 0-577 4 = 57-38 m 


AC = 


DC = 99-38 




BC = AC - AB 




= 99-38 - 42 = 57-38 m 



Check 



Exercises 2(a) 

1. A flagstaff 90 ft high is held up by ropes, each being attached to 
the top of the flagstaff and to a peg in the ground and inclined at 30° 
to the vertical; find the lengths of the ropes and the distances of the 
pegs from the foot of the flagstaff. 

(Ans. 103-92 ft, 51-96 ft) 

2. From the top of a mast of a ship 75 ft high the angle of depression 
of an object is 20°. Find the distance of the object from the ship. 

(Ans. 206 -06ft) 

3. A tower has an elevation 60° from a point due north of it and 45° 
from a point due south. If the two points are 200 metres apart, find the 
height of the tower and its distance from each point of observation. 

(Ans. 126-8 m, 73-2 m, 126-8 m) 

4. A boat is 1500 ft from the foot of a vertical cliff. To the top of the 
cliff and the top of a building standing on the edge of the cliff, angles 
of elevation were observed as 30° and 33° respectively. Find the 
height of the building to the nearest foot. 

(Ans. 108 ft) 

5. A vertical stick 3 m long casts a shadow from the sun of 1-75 m. 
What is the elevation of the sun ? 

(Ans. 59° 45') 

6. X and Y start walking in directions N17°W and N73°E; find their 
distance apart after three hours and the direction of the line joining 
them. X walks at 3 km an hour and Y at 4 km an hour. 

(Ans. 15 km S70°08'E) 

7. A, B, and C are three places. B is 30 km N67H°E of A, and C 



72 SURVEYING PROBLEMS AND SOLUTIONS 

is 40 km S22 1 / 2 °E of B. Find the distance and bearing of C from A. 

(Ans. 50 km, S59°22'E) 

2.2 Circular Measure 

The circumference of a circle = 2rrr where 77 = 3*1416 approx. 

2.21 The radian 

The angle subtended at the centre of a circle by an arc equal in 
length to the radius is known as a radian. 

Thus 2 77 radians = 360° 

/. 1 radian = M 

277 

= 57°17 , 45" approx. 

= 206 265 seconds. 

This last constant factor is of vital importance to small angle cal- 
culations and for conversion of degrees to radians. 

Example 2.5. Convert 64° 11' 33" to radians. 

64° 11' 33" = 231093 seconds. 

.-. no. of radians = 231093 = 1-120 37 rad. 
206 265 

Tables of radian measure are available for 0°-90° and, as the 
radian measure is directly proportional to the angle, any combination 
of values produces the same answer for any angular amount. 

By tables, 



64° 






= 


1-11701 




11' 




= 


0-003 20 






33" 


= 


0-000 16 


64° 


11' 


33" 


— 


1-120 37 



It now follows that the length of an arc of a circle of radius r and 
subtending d radians at the centre of the circle can be written as 

arc = r.6 rad (2.14) 

This is generally superior to the use of the formula 

■* - 2m * m (21S) 

N.B. When 6 is written it implies 6 radians. 
To find the area of a circle. 

A regular polygon ABC ... A is drawn inside a circle, Fig. 2.17. 



SURVEYING TRIGONOMETRY 



73 



Draw OX perpendicular to AB 
Then area of polygon = 

±OX(AB + BC + ...) = 
■tOX (perimeter of polygon) 

When the number of sides of the 
polygon is increased to infinity 
(oo), OX becomes the radius, 
the perimeter becomes the cir- 
cumference, and 
the polygon becomes the circle 

.*. area of circle = V2.r .2ttt 




= TFT 2 




The area of the sector OAB. 




area of sector 


e 


area of circle 


2tt 


ari=>a of sftrtnr ^ rrr " _ 

2tt 


\r*Q 



Fig. 2. 17 



(2.16) 



2.22 Small angles and approximations 

For any angle < 90° (i.e. < 77/2 radians) tantf > > sim 

A 




Fig. 2. 18 
Let angle AOC = 6 

OA = OC = r 
and let AB be a tangent to the arc AC at A to cut OC produced at B. 
Draw AD perpendicular (1) to OB. 



74 SURVEYING PROBLEMS AND SOLUTIONS 

Then area of triangle OAB = ±0A . AB 

= h.rtand = ±r 2 tan0 

2 2 

area of sector OAC = |r 2 
area of triangle OAC = ±OC.AD 

= Ir.r sin0 = |r 2 sin0 



Now triangle OAB > sector (X4C > triangle OAC. 
Ar 2 tan0 > |r 2 > |r 2 sin0 
tan0 > 9 > sin0 

This is obviously true for all values of < it/2. 

Take to be very small. 
Divide each term by sin0, 

then * > , 3 > 1 

cos0 sin0 

It is known that as -* then cos0 -* 1. 

Thus cos0 ~ 1 when is small 

.*. .. must also be nearly 1. 

sin0 

The result shows that sin0 may be replaced by 0. 
Similarly, dividing each term by tan0, 

1 > A > s * n fl 
tan0 tan0 

i.e. 1 > " > cos0 

tan0 

Tan0 may also be replaced by 6. 

It can thus be shown that for very small angles 
tan0 =2.. ^ sin0. 

The following values are taken from H.M. Nautical Almanac Office. 
Five-figure Tables of Natural Trigonometrical Functions. (These 
tables are very suitable for most machine calculations.) 



SURVEYING TRIGONOMETRY 



75 



Angle 


Tangent 


Radian 


Sine 


1°00'00" 


0-01746 


0-01745 


0-01745 


1° 30' 00" 


0-026 19 


0-02618 


0-026 18 


2° 00' 00" 


0-03492 


0-03491 


0-03490 


2° 30' 00" 


0-04366 


0-04363 


0-04362 


3° 00' 00" 


0-05241 


0-05236 


0-05234 


3° 30' 00" 


0-061 16 


0-06109 


0-06105 


4° 00' 00" 


0-06993 


0-06981 


0-06976 


4° 30' 00" 


0-07870 


0-078 54 


0-078 46 


5° 00 '00" 


0-08749 


0-08727 


0-08716 



From these it can be seen that $ may be substituted for sin 6 or 
tan0 to 5 figures up to 2°, whilst 6 may be substituted for sin0 up 
to 5° and for tantf up to 4° to 4 figures, thus allowing approxima- 
tions to be made when angles are less than 4° 



Example 2.6. If the distance from the earth to the moon be 250000 
miles (402000 km) and the angle subtended 0° 30', find the diameter of 
the moon. 



250 OOO miles 




Fig. 2. 19 
The diameter ~ arc ABC 

2L 250000 x 30'rad 

- 250000 v 30 x 60 
206265 



=2 2180 miles (3510 km) 



Example 2.7. Find as exactly as possible from Chambers Mathemati- 
cal Tables the logarithmic sines of the following angles: 

A = 00° 02' 42" and C = 00° 11' 30" 

Use these values to find the lengths of the sides AB and AC in a tri- 
angle ABC when BC = 12-736 ft. Thereafter check your answer by 
another method, avoiding as far as possible using the tables at the 
same places as in the first method. 

The lengths are to be stated to three places of decimals. 

(M.Q.B/S) 



76 SURVEYING PROBLEMS AND SOLUTIONS 

As the sines and tangents of small angles change so rapidly, 
special methods are necessary. 

Method 1 

Chambers Mathematical Tables give the following method of finding 
the logarithmic sine of a small arc: 

To the logarithm of the arc reduced to seconds, add 4-685 574 9 and 
from the sum subtract 1/3 of its logarithmic secant, the index of 
the latter logarithm being previously diminished by 10. 

00°02'42" = 162" logl62 2-2095150 

constant 4-6855749 

-|(logsec02'42" - 10) = | x 0-000 000 2 
00° 11" 30" = 690" 



-!(log sec 11' 30" - 10) = I x 0-0000024 



Method 2 

As the sines and tangents of small angles approximate to the 
value, 

radian value of A 00° 02' 42" = * 6 ^ = 0-000785 3974 

206 265 

logO-000785397 4 = 6-8950895 

radian value of B 00° 11' 30" = „ 690 „ = 0-003 345 2112 

206 265 

logO-003 3452112 = 7-524423 5 

(2) Vegas tables (to l") 

6-895089 5 6-8950898 

7-524 4235 7-524 4231 



To find the length of sides AB and AC when BC = 12-736 : 

AB = BC sin C = 12-736 sin (02' 42" + 11' 30") (see 2.51 
sin A sin 02' 42" 





6-8950899 


= 


0-0000001 




6-8950898 


log 690 


2-8388491 


constant 


4-685 5749 




7-5244240 


= 


0-0000008 




7-524423 2 



Summary 


(1) 


00° 02' 42" 


6-895089 8 


00° 11' 30" 


7-524 423 2 



SURVEYING TRIGONOMETRY 



77 



AB = 66-803 ft 



AC = 



BC sin B 
sin^4 



AC = 54-246 ft 





Logs 


12-736 


= 1-1050331 


S/14' 12" 


= 2-929 2793 




4-6855749 




8-719887 3 


-|(sec-10) 


0-0000035 




8-7198838 


S/02' 42" 


6-8950898 


AB 


1-8247940 


12-736 sin 11' 30" 




sin 02' 42" 




12-736 


1-1050331 


S/ll' 30" 


7-524423 2 




8-629 4563 


S/02' 42" 


6-895 0898 


AC 


1-7343665 



2.3 



Trigonometrical Ratios of the Sums and Differences 
of two angles (Fig. 2.20) 
To prove: 

sin (A + B) = sin A cos B + 

+ cos A sin B 

(2.17) 

cos (A + B) = cos A cos B - 

- sin A sinS 

(2.18) 

Let the line OX trace out the 
angle A and then the angle B. 
Take a point P on the final line 
0X 2 . Draw PS and PQ perpendicu- 
lar to OX and OX, respectively. Fig. 2.20 
Through Q draw QK parallel to OX to meet PS at R. Draw QT per- 
pendicular to OX. 

ffPQ = KQO = A 
RS + PR 




sin(A + B) = fl = 
OP 



OP 



RS , PR 
OP OP 



= RS OQ PR PQ 
OQ 'OP PQ' OP 
= sin^4 cos 6 + cos A sinB 



78 SURVEYING PROBLEMS AND SOLUTIONS 

cosC4 + fi) = OS = OT-ST m OT_ST 
OP OP OP OP 

OT 0£_ ST P£ 
OQ' OP PQ' OP 

= cos 4 cosB - sin 4 sinB 

If angle B is now considered -ve, 

sin(4-B) = sin 4 cos (-B) + cos 4 sin (-B) 

= sin 4 cosB - cos 4 sinB (2.19) 

Similarly, 

cos(4-B) = cos 4 cps(-B) - sin 4 sin(-B) 

= cos A cos B + sin A sin B (2.20) 

tanG4'+B) = s * n (A + ff) _ sin A cosB + cos A sinB 
cos (4 + B) cos A cos B - sin4 sin B 

(-by cos4 cosB) = tan 4 + tanB (2 . 2 1) 

1 - tan A tan B 

Similarly, letting B be -ve, 

tanC4-B) = tan 4 + tan(-B) 
l-tan4 tan(-B) 

tan A - tanB 



1 + tan4tanB (2.22) 

If sin(4 + B) = sin A cosB + cos 4 sinB 

and sin (A - B) = sin A cosB - cos A sinB 

then sin (A + B) + sin(4 - B) = 2 sin A cosB (2.23) 

and sin(4 + B) - sin (A - B) = 2 cos A sinB (2.24) 

Similarly, 

as cos (4 + B) = cos 4 cosB - sin 4 sinB 

and cos(4 - B) =• cos 4 cos B + sin 4 sin B 

cos(4 + B) + cos(4 - B) = 2 cos 4 cosB (2.25) 
cos(4 + B) - cos(4 - B) = -2 sin 4 sinB (2.26) 

If A « B, then 

sin (4 + 4) = sin 2A = 2 sin4 cosA (2.27) 

cos(4 + 4) = cos 24 = cos 2 4 - sin 2 4 

= 1 - 2 sin 2 4 (2.28) 

or =2 cos 2 4 - 1 (2.29) 



SURVEYING TRIGONOMETRY 

2 tan A 



tan04 + 4) = tan 2 A = 



1 - tan 2 A 



From 



if 
and 

then 



Similarly, 



From 



2.4 Transformation of Products and Sums 

sin(4 + B) + sin(A - B) = 2 sin A cosB 
sin 04 + B) - sinC4 - B) = 2 sinB cos 4 

A + B = C 

A - B = D 
A = C-±D and B = £^_D 



cos(v4 + B) + cos(A - B) = 2 cos ,4 cosB 
cos04 + B) - cos(A - B) = -2 sinA sinfi 

C - D 



cos C + cos D = 2 cos c + D 



cos 



and cosC - cosD = -2 sin C + D sin C ~ P 
2 2__ 

These relationships may thus be tabulated : 

sin (A ± B) = sin A cos B ± cos A sin B 
cos 04 ± B) = cos /4 cos B + sin A sin B 



tan 04 ±B) 



tan>l ± tanB 



79 
(2.30) 



2 "" " 2 
sin C + sinD = 2 sin C + D cos C ~ D 


(2.31) 


sinC sinD - 2 sin 6 " ~ D cos C + D 

2 2 


(2.32) 



(2.33) 
(2.34) 



1 T tan 4 tan B 
sin04 + B) + sin 04 - B) = 2 sin A cosB 
sin04 + B) - sin (4 - B) = 2 cos A sinB 
cos(i4 + B) + cos(A - B) = 2 cos 4 cos B 
cos(/4 + B) - cos(/4 -B) = -2sini4sinB 

sin 2,4 = 2 sin ,4 cos /} 

cos 2A = cos 2 A - sin 2 A = 1 - 2sin 2 4 = 2 cos 2 ,4 - 1 

tan 24 = 2tan ^ 



1 - tanM 
sin A + sinB 



2 sin 



4 + B„^A - B 



cos 



80 SURVEYING PROBLEMS AND SOLUTIONS 

sin A - sin B = 2 sin' 4 ~ B cos^ 4 + B 

2 2 

cos 4 + cos £ = 2 cos ^ + B cos ^ ~ g 

2 2 

cos ,4 - cos B = -2 sin ^ + B sin ^ ~ B 

2 2 

2.5 The Solution of Triangles 

The following important formulae are now proved: 

Sine rule 

. a = b = _£_ = 2R (2.35) 

sin A sinS sinC 

Cosine rule 

c z = a 2 + b z - 2ab cosC (2.36) 



sinC = 2l^\s(s -a){s - b){s - c)\ (2.37) 

ab 

Area of triangle = lab sinC (2.38) 



= ^s(s-a)(s-b)(s-c) (2.39) 

Half-angle formulae 

sini = f (s-b)(s-c) (2 .40) 

2 V be 



tan 

Napier's tangent rule 

B -C 
an — — -l = 

b + c 



A = A(s-a) (2.41) 

2 W be 

A = As-b)(s-c) (2.42) 

2 V s ( s _ a) 



tan*^ = An£tan£±^ ( 2>43) 



2.51 Sine rule (Figs. 2.21 and 2.22) 

Let triangle /4SC be drawn with circumscribing circle. 

Let 4J3, be a diameter through /I (angle ABC = angle 4^0. 

AC 
In Fig. 2.21, j£- = sinB 

In Fig. 2.22, i£ = sin(180-B) 

AB | 

= sinB 

JL = sinB 
2R 



SURVEYING TRIGONOMETRY 




81 




Fig. 2.21 



Fig. 2.22 



Similarly 



sin A 



b 
sinB 

b 
sinB 



= 2R 



sinC 



= 2R 



(2.35) 



2.52 Cosine rule (Fig. 2.23) 




AB 2 = 



Obtuse 

Fig. 2. 23 The cosine rule 
.2 „„ 2 



Acute' 



with C acute 



AD 2 + BD 2 (Pythagoras) 
= AD 2 + (BC - CD) 2 
= b 2 sin 2 C + {BC - b cos Cf 
= AD 2 + (BC + CD) 2 

= 6 2 sin 2 (180 - O + {BC + b cos (180 - C)\ 2 } with C obtuse 
= b 2 sin 2 C + (BC - b cos C) 2 
.'. AB 2 = b 2 sin z C + (BC-bcosC) 2 in either case 



= b 2 sin 2 C + a z - 2ab cosC + b z cos z C. 
= a 2 + b 2 (sin 2 C + cos 2 C) - lab cosC 
= a 2 + b 2 - lab cos C 



(2.36) 



82 SURVEYING PROBLEMS AND SOLUTIONS 

2.53 Area of a triangle 

From c 2 = a 2 + b 2 - lab cosC 

cos C = 



~2 u z ~ z 
a + b - c 



sin 2 C = 1 - cos 2 C = 1 _( a Z +b 2 -c 2 \ 2 

\ lab / 



lab 

lab 



= (l + a z +b 2 -c% a 2 + b 2 -c 2 ) 
V lab A lab J 

_ (a + b) 2 - c 2 x c 2 - (a - b) 2 



2ab 2afe 

(a + b + c)(- c + a + bXc - a + b)(c + a - fc) 
(2ab) 2 

4s (s - a) (s - b)(s - c) 



a 2 b z 
where 2s = a + b + c. 



ab*j 



sinC = i_ /s(s-a)(s-b)(s-c) (2.37) 



In Fig. 2.23, 



Area of triangle = \AD . EC 



= lab sin C (2.38) 



= ^Lab -4 V s ( s - a)(s - b)(s - c) 



= ^s(s-a)(s-b)(s-c) (2.39) 

2.54 Half-angle formulae 
From Eq.(2.28), 

tint* A -%1-co.A) . ^(l-»Lt£L-L) 

q 2 _ (fc _ C ) 2 
4bC 

(a - b + c)(a + b - c) 
Abe 

= (s-b)(s-c) 
be 

sini = Ks-bKs-e) (2 40) 

2 V fee 



SURVEYING TRIGONOMETRY 83 

Similarly, 

cosz±A = I(l + cos4) = l(l + b 2 +c 2 -a 2 \ 
2 2 2\ 2bc * 

= (b + c) 2 - a 2 
4bc 

= (b + c + a)(b + c - a) 
Abe 

_ s(s - a) 
be 

cosil = /sjs-a) (241) 

2 V be 



tan_d = 



. A 
sin 



cos 
2 



- = / (s-bXs-c) (2>4 2) 

A V s(s - a) 



The last formula is preferred as (s - b) + (s - c) + (s - a) = s, which 
provides an arithmetical check. 

2.55 Napier' s tangent rule 

From the sine rule, 



then 





b _ sinB 
c sinC 


b - c 


sinS - sinC 


b + c 


sinS + sinC 




2cos S + C sin B - C 
2 2 




r sin B + C ™c B-C 

2 2 




tan B ~ C 
2 




B + C 
tan 



(By Eqs. 2.31/2.32) 



•*. tan B ~ c = b-c tQV B + C (2.43) 

2 b + c 2 

2.56 Problems involving the solution of triangles 

All problems come within the following four cases: 



84 



SURVEYING PROBLEMS AND SOLUTIONS 



(1) Given two sides and one angle (not included) to find the other 
angles. 

Solution : Sine rule solution ambiguous as illustrated in Fig. 
2.24. 



Given AB and AC with angle B, 
AC may cut line BC at C, or C 2 




Fig. 2.24 The ambiguous case of the 
sine rule 

(2) Given all the angles and one side to find all the other sides. 
Solution: Sine rule 

(3) Given two sides and the included angle 

Solution: Either cos rule to find remaining side 

or Napier's tangent rule (this is generally pre- 
ferred using logs) 

(4) Given the three sides 
Solution: either cos rule 

or half-angle formula (this is generally preferred 
using logs.) 



Example 2.8 (Problem 1) 



Let 


c = 


466-0 m 




a = 


190-5 m 




A = 


22° 15' 


Using sine rule, 






sin C sin A 






c a 




sinC = csinA 
a 







= 466-0 sin 22° 15' 
190-5 

466-0 x 0-378 65 
190-5 

= 0-926 25 




Fig. 2.25 



SURVEYING TRIGONOMETRY 85 



Angle C 2 = 67° 51' 30" or 180 
C, = 112 08' 30" 

To find side b (this is now Problem 2) , 



a 



sin B sin A 

fo _ a sinB 
sin ,4 



i.e. 



5, = 



Log calculation 



190-5 sin [180 -(67° 51' 30" +22° 15' 00")] 
sin 22° 15' 00" 

190-5 sin89°53'30" 
sin 22° 15' 00" 

= 190 ' 5 x 1-0 = 503-10 m 
0-37865 

190-5 sin[(180 -(112° 08' 30"+ 22° 15' 00")] 
sin 22° 15 '00" 

190-5 sin 45° 36' 3 0" = 359 . 51m 
sin 22° 15' 00 



log sin C = logc + log sin A - log a 
C = 67° 51' 30" 

466-0 ©2-668 39 

sin 22° 15' 9-578 24 

©2-24663 

190-5 2-279 90 

sinC ©9-96673 

N.B. The notation 9-578 24 is preferred to 1-578 24 - this is the form 
used in Chambers, Vegas, and Shortredes Tables. 

Every characteristic is increased by 10 so that subtraction is sim- 
plified — the ringed figures are not usually entered. 

Also logfc = log a + log sin B + log cosec,4 

"I.B. Addition using log cosec4 is preferable to subtracting log sin A. 
e. logfc, = logl90-5 + log sin 89° 53' 30" + log cosec22° 15'00" 

190-5 2-27990 

sin89°53'30' 0-0 

cosec22°15'00" 10-42176 

fr, = 503-10 m b, 2-70166 



86 SURVEYING PROBLEMS AND SOLUTIONS 

log& 2 = logl90-5 + log sin 45° 36' 30 " + log cosec22° 15'00" 

190-5 2-27990 

sin45°36'30" 9-85405 

cosec 22° 15' 00" 10-421 76 

b z = 359-51 m b 2 2-55571 

N.B. A gap is left between the third and fourth figures of the logarithms 
to help in the addition process, or it is still better to use squared paper. 



Example 2.9 (Problem 3) 


Let 


a = 636 m 




c = 818 m 




B = 97° 30' 


To find b, 


A and C. 


By cosine rule 




b 2 = 


a* + c 2 - 2ac cosB 


= 


636 2 + 818 2 - 2 x 636 x 818 x cos 


= 


404496 + 669124 + 135815-94 


= 


1209435-94 


b . 


1099-74 m 




sin A sin B 




a b 




oinA asinB 636 sin 97° 30' 
SmA = b = 1099-74 




= 0-57337 




A = 34° 59' 10" 



B = 47° 30' 50" 

The first part of the calculation is essentially simple but as the 
figures get large it becomes more difficult to apply and logs are not 
suitable. The following approach is therefore recommended. 

As c> a, C> A. Then, by Eq.(2.43), 

tan%4 , £^JltanC±J. 
2 c + a 2 

818 - 636 . (180 -97° 30') 
818 + 636 2 

182 tan 41° 15' 
1454 

= 0-10977 



c 


= 


47° 30' 


50" 




A 


= 


34° 59' 


10" 




b 


= 


a sin 6 


cosec 


A 



SURVEYING TRIGONOMETRY 87 

\{C-A) = 6° 15' 50" 

l -(C + A) = 41° 15' 00" 

By adding 
By subtracting 

Now, by sine rule, 

= 636 sin 97° 30' cosec 34° 59' 10" 

= 1099-74 m 
N.B. This solution is fully logarithmic and thus generally preferred. 
Also it does not require the extraction of a square root and is there- 
fore superior for machine calculation. 

Example 2.10 (Problem 4) 

Let a = 381 b = 719 c = 932 

To find the angles. 

a 2 

cos A 



From 
then 



b 2 + c 2 - 2bc cos A 

b 2 + c 2 - a 2 
2bc 

719 2 + 932 2 - 381 2 

2 x 719 x 932 

516 961 + 868 624 - 145 161 



1 340 216 



A = 



0-925 54 
22° 15' 00" 



cosfi 



2 „2 t 2 

a + c - b 
2ac 

145161 + 868 624 - 516961 



2 x 381 x 932 



496824 



710 184 
B = 45° 36' 30 



= 0-69957 



cosC = 



a 2 + b 2 - c 2 
lab 

145161 + 516961 - 868624 

2 x 381 x 719 
-206 502 



547 878 



= -0-37691 



88 SURVEYING PROBLEMS AND SOLUTIONS 

C = 180 - 67° 51' 30" 
= 112° 08' 30" 

Check 22° 15' 00" + 45° 36' 30" + 112° 08' 30" = 180° 00' 00" 
Alternative 

By half-angle formula, tan S. = Ks - b)(s - c) 

2 V s(s - a) 

a = 381 s - a = 635 

b = 719 s - b = 297 

c = 932 s - c = 84 

2s =2032 



s = 1016 s = 1016 

then tanA - / 297 * 84 

2 V 1016 x 635 

This is best solved by logs 

log tani = I[(log297 + log84) - (log 1016 + log635)] 



2 2 

= 11° 7' 30" 
= 22° 15' 00" 



297 2-47276 

84 1-924 28 



tan^ = / 635 >< 84 
2 V1016 x 297 



-^ = 22° 48' 15" 
2 

B = 45° 36' 30" 







4-39704 


1016 




3-00689 


635 


2) 

1/2 


2-80277 

5-80966 

18-58738 


tany 


9-29369 


635 




2-80277 


84 




1-924 28 
4-727 05 


1016 




3-00689 


297 




2-47276 
5-47965 




2) 19-24740 


tan B/2 


9-623 70 



Example 2.11 



SURVEYING TRIGONOMETRY 89 

The sides of a triangle ABC measure as follows: 



AB = 36ft Olin., AC = 30 ft if in. and BC = 6ft0lin. 

lo 4 4 

(a) Calculate to the nearest 20 seconds, the angle BAC. 

(b) Assuming that the probable error in measuring any of the sides 
is ± 1/32 in. give an estimate of the probable error in the angle A. 

(M.Q.B/S) 




Fig. 2. 26 
AB = c = 36 ft 0* in. = 36-036 ft 

16 

AC = b = 30ftlf-in. = 30-146 ft 

4 



S - C = 



BC = a = 



Using Eq. (2.42) 



6ft0±in. = 

4 



6-021 ft 



2 |72-203 
s = 36-1015 



c = 0-0655 
5-9555 

s - a = 30-0805 
s = 36-1015 



tan-d = /Cs-bHs-c) 
2 V s(s-a) 



J, 



5-955 5x0-065 5 



36*1015x30-0805 

4- = 1°05'10" 
2 

A = 2 o 10 , 20 /, 

tan# = / (s-o)(s-"c) 
2 V s(s - 6) 



y 



30-0805x0-065 5 
36-1015x5-9555 



^ = 5° 28' 05" 
2 

g = 10° 56 '10" 

tan£ = Rs-a)(s-b) 
2 V s(s-c) 



■J 



30-0805x5-9555 
36-1015x0-0655 



90 SURVEYING PROBLEMS AND SOLUTIONS 

£. = 83° 26' 45" 
2 

C = 166° 53' 30" 

Check A + B + C = 180° 

(b) The probable error of ±1/32 in. = ±0*003 ft. 

The effect on the angle A of varying the three sides is best cal- 
culated by varying each of the sides in turn whilst the remaining two 
sides are held constant. To carry out this process, the equation must 
be successively differentiated and a better equation for this purpose 
is the cosine rule. 

b z + c 2 - a 2 



Thus cos A = 

Differentiating with respect to a, 

-sin A8A a = 

8A„ = 



2bc 



2a 8a 
2bc 

a 8a 
be sin A 



Differentiating with respect to b, 

-sin A8A h = (2bcx2b)-(b 2 +C 2-a*)(2c) 



4b*c'- 



but cos C 



X 

c 2b 2 c 

a 2 + b - c 2 
2b 2 c 

a + b - c 



2ab 

_ :osC 

* 6 be sin A 



8A b = - a cos C 8b = - 8A a cos C (as 8a = 8b) 



Similarly, from the symmetry of the function: 

8A C = _ « cos g g c = SA a cos B (as 8a = 8c) 
be sin A 

Substituting values into the equations gives: 

8A = 6-021 x ±0-003 x 206265 ±90-49 sec 

30-146 x 36-036 x sin 2° 10' 20" 

8A b = S4 cos 166° 53' 30" = ±88-13 sec 

8A„ = &4„cos 10° 56' 10" = ±88-85 sec 



SURVEYING TRIGONOMETRY 



91 



Total probable error = V&4 2 . + 5-4 2 + 8A 2 C 

= V90-49 2 + 88-13 2 + 88-85 : 
= ± 154 seconds. 



2.6 Heights and Distances 

2.61 To find the height of an object having a vertical face 

The ground may be (a) level or (b) sloping up or down from the 
observer, 
(a) Level ground (Fig. 2.27) 



Fig. 2.27 




V.D. 



The observer of height h is a horizontal distance (H.D.) away 
from the object. The vertical angle (V.A.) = 6 is measured. The 
vertical difference 

V.D. = H.D. tanfl (2.44) 

Height of the object above the ground = V.D. + h 



V.D. 



V.D. 




\B Horizontal line 

._o< Depression" 




H.D 




Fig. 2-28 



(b) Sloping ground (Fig. 2.28) 

The ground slope is measured as a 

V.D. = H.D. (tan0 ± tana) (2.45) 

Height of object above the ground = V.D. + ft 

N.B. This assumes that the horizontal distance can be measured. 



92 



SURVEYING PROBLEMS AND SOLUTIONS 



2.62 To find the height of an object when its base is inaccessible 

A base line must be measured and angles are measured from its 
extremities. 

(a) Base line AB level and in line with object. (Fig. 2.29) 
Vertical angles a and /8 are measured. 




Fig. 2 29 
A,C, = EC, cot a 
B,C, = EC, cot/8 
A,B, = A,C, - B,C, = EC, (cot a -cot 0) 



Thus 



EC, = 



AB 



cot a - cot j8 
Height of object above ground at A 

= EC, + ft, 
Height of ground at D above ground at A 

= EC, + h, — h 2 



(2.46) 



(b) Base line AB level but not in line with object (Fig. 2.30) 

Angles measured at A horizontal angle 

vertical angle a 

at B horizontal angle <fi 

vertical angle /8 



In triangle ABC ; 



AC = AB sin<£ cosec(0 + 0) (sine rule) 



SURVEYING TRIGONOMETRY 



93 




Fig. 2.30 
and BC = AB sin 6 cosec(0 + <£) 

Then C,£ = 4Ctana 

= ,46 sin cosec(0 + <£) tana (2.47) 

Also C,£ = BC tan/8 

= 4B sin0 cosec(0 + 0) tan£ (2.48) 

Height of object (E) above ground at A 

= C,£ + A, 
Height of ground (D) above ground at A 

= C,E + /i, - /i 2 
(c) Sase Zme 4B on sloping ground and in line with object (Fig. 2.31) 




Fig. 2.31 



94 SURVEYING PROBLEMS AND SOLUTIONS 

Angles measured at A: vertical angle a (to object) 

vertical angle 8 (slope of ground) 
Angle measured at B : vertical angle j8 (to object) 
In triangle A^EB X , 

4, = a - 8 

B, = 180 - (£ - 8) 

E = p - a 

Then A,E = A,B, sin {180 -(0-S)} cosec(/8 - a) 

= AB sin(/8 - 8) cosecQ8 - a) (2.49) 

Height of object (£) above ground at A 
EC = EC, + h } 

= A^E sin a + /i, 

= AB sin(fi - 8) cosec(/8 - a) sina + /i, (2.50) 

Height of ground (D) above ground at A 

= EC, + hy — h 2 . 
(d) itase line AB on sloping ground and not in line with object (Fig.2.32) 




Fig. 2.32 

Angles measured at A : horizontal angle 6 

vertical angles a (to object) 

8 (slope of ground) 
at B: horizontal angle <f> 
vertical angle /S 

A y B 2 = AB cos 8 



SURVEYING TRIGONOMETRY 



95 



Then 4,0, = A X B 2 sin<£cosec(0 + <f>) 

EC, = 4,C, tana 

— AB cos 8 sin0 cosec(fl + <ft) tana (2.51) 

Height of object (E) above ground at A 

EC = EC X + hi 
E,C 2 = B 2 C, = y4,B 2 sin0 cosec(0 + <£) 
EC 2 = B,C 2 tan/3 
Similarly, height of object (E) above ground at j4 
EC = EC 2 + fi,B 2 + h, 

= 4g cos 5 sing cosec(fl + <f>) tan/6 + AB sing + ft, ,~ co\ 
Height of ground (D) above ground at 4 

= EC\ + hi — h z 
or = EC 2 + h, - h 2 + AB sinS 

2.63 To find the height of an object above the ground when its base 
and top are visible but not accessible 

(a) Base line AB, horizontal and in line with object (Fig. 2.23) 
Vertical angles measured at A : a, and a 2 

at B: /S, and B z 




Fig. 2.33 



96 SURVEYING PROBLEMS AND SOLUTIONS 

From Eq.(2.46) 



Also 



C,E = 
C,D = 



AB 



cot a, - cot/3, 
AB 



cot a 2 - cotB z 
Then ED = C,E - C,D = H 



H = AB\ 



[cota, - cot^3, cota 2 - cotB z J 

(b) Base line AB horizontal but not in line with object (Fig. 2.34) 
Angles measured at A : horizontal angle 6 

vertical angles a, and a 2 
at B : horizontal angle <f> 

vertical angles jS, and B z 



(2.53) 




I MHEWM^WAW/SigWig 



Fig. 2. 34 



AC 


= 4,C, 


ED 


= H 


.'. 


H 


Similarly, 






H 



= AB sin0 cosec(# + <f>) 

= AC (tan a, - tan a 2 ) 

= AB sin <f> cosec(0 + ^>)(tana, - tana 2 ) (2.54) 

= AB sin 6 cosec (0 + <£) (tan B , - tan /8 2 ) (2 . 55) 



SURVEYING TRIGONOMETRY 



97 



(c) Base line AB on sloping ground and in line with object (Fig. 2.35) 
Vertical angles measured at A : a,, a 2 and 8 

at B: /8, and /8 2 




Fig. 2.35 
From Eq.(2.50) 

EC, = /IB sin(/3, - 8) cosec(/9, - a,) sin a, 
and DC, = 4B sin(/3 2 - 5) cosecQ3 2 - a 2 ) sin a 2 

Then ED = EC, - DC, 

H = A8[sin(/6, - 5) cosec(/3, - a,) sin a, - 

- sin(/8 2 - 8) cosec(j8 2 - a 2 ) sin a 2 ] (2.56) 

(d) Base line AB on sloping ground and not in line with object 
(Fig. 2.36) 

Angles measured at A: horizontal angle 6 

vertical angles a, and a 2 

8 (slope of ground) 
at B: horizontal angle cf> 

vertical angles /3, and £ 2 



98 



SURVEYING PROBLEMS AND SOLUTIONS 




Fig. 2. 36 
From Eq.(2.51), 

ECi = AB cosS sin<£ cosec(0 + <f>) tana, 
DC X = AB cos 8 sin<£ cosec(<9 + <f>) tana 2 
Then ED = EC, - DC, 

= AB cos 5 sin0 cosec(0 + <£)(tana, - tana 2 ) (2.57) 
Also ED = AB cos 5 sin 6 cosec(<9 + <£)(tan)6, - tan B z ) (2.58) 



2.64 To find the length of an inclined object (e.g. an inclined 
flagstaff) on the top of a building (Fig. 2.37) 

Base line AB is measured and if on sloping ground reduced to 
horizontal. 

Angles measured at A: horizontal angles 0, and d z to top and 

bottom of pole 
vertical angles a, and a 2 to top and 

bottom of pole 
at B: horizontal angles <£, and <f> 2 to top and 

bottom of pole 
vertical angles )6, and B z to top and 

bottom of pole. 
In plan the length ED is projected as E^D^ (= E Z D). 

In elevation the length ED is projected as EE 2 , i.e. the difference in 
height 

Then AE, = AB sin</>, cosec(0, + <£,) 

Also AD, = AB sin 2 cosec(0 2 + </> 2 ) 



SURVEYING TRIGONOMETRY 



99 




e d 



Fig. 2.37 

The length E,D = L is now best calculated using co-ordinates 
(see Chapter 3) 

Assuming bearing AB = 180° 00' 

Ae = AE, sin(90 -6,) = AE, cos0, 
4d = 4D, sin(90 - 2 ) = 4D, cos0 2 

Then ed = D 2 D, = /Id - 4e 

= AD, cos6 2 - 4E, cos0, 

and similarly, E,D 2 = E,e _ D 2 e = 4£, sin0, - AD, sin0 2 
In the triangle E,D,D 2 . 

The bearing of the direction of _ tan -i D 2 D, 
inclination (relative to AB) E,D 2 

Length E,D, = E,D 2 (sec bearing) 

To find difference in height E E 2 

Height of top above A = AE, tana, 

Height of base above A = AD, tan a 2 

Length EE 2 = AE, tana, - AD, tan a 2 



To find length of pole : 
In triangle EDE 2 , 



i.e. 



ED 2 
ED 



EE\ + E 2 D Z 



-V 



EEl + E,D 2 



100 



SURVEYING PROBLEMS AND SOLUTIONS 



2.65 To find the height of an object from three angles of elevation 
only (Fig. 2.38) 




Fig. 2.38 
Solving triangles ADB and ADC by the cosine rule, 

cos* = h 2 cot 2 a + x z -h 2 cot z p 
2hx cot a 

fc 2 cot 2 a + (x + y) 2 - h 2 cot 2 



(2.59) 



2h(x + y) cot a 
••• (x+y)[fc 2 (cot 2 a - cot 2 j8) + x 2 ] = x[h\cot 2 a - cot 2 0) + (x + y) 2 ] 

i.e. h 2 [(x + y)(cot 2 a - cot 2 j9) - x(cot 2 a - cot 2 0)] 

= x(x + y) 2 - x 2 (x + y) 



i.e. h 2 = 



(x + y)[x(x + y) - x 2 ] 



(x + y)(cot 2 a - cot 2 /S) - x(cot 2 a - cot 2 0) 

(x + y)(xy) 

(x + y)(cot 2 a - cot 2 j8) - x(cot 2 a - cot 2 0) 

[ xy(x + y) ] 

[x(cot 2 - cot 2 /8) + y(cot 2 a - cot 2 j8)J 



H x = y, 



h = 



y/2x 



[cot 2 6 - 2 cot 2 /3 + cot 2 a]* 



(2.60) 



(2.61) 



SURVEYING TRIGONOMETRY 



101 



Example 2.12 A,B and C are stations on a straight level line of 
bearing 126° 03' 34". The distance AB is 523-54 ft and BC is 420-97 
ft. With an instrument of constant height 4'- 3" vertical angles were 
successively measured to an inaccessible up-station D as follows: 

At A 7° 14' 00" 

B 10° 15' 20" 

C 13° 12' 30" 
Calculate (a) the height of station D above the line ABC 

(b) the bearing of the line AD 

(c) the horizontal length AD. 

(R.I.C.S.) 
D 




Fig. 2.39 

(a) In Fig. 2.39, 

AD = h cot a 

BD = h cot/8 

CD = h cot0 

Solving triangles AD % B and AD X C, using Eq. (2.60), 

_ i 

1 2 



h = 



xy(x + y) 



*(cot 2 - cot 2 /9) + y(cot 2 a - cot 2 /S) 

(523-54 x 420-97)(523-54 + 420-97) 



523-54(cot 2 13° 12' 30 - cot 2 10° 15' 20") + 420-97(cot 2 7° 14' 00" - cot 2 10 
= 175-16 ft 



— 1 

'15'20"J 



102 



SURVEYING PROBLEMS AND SOLUTIONS 



.*• Difference in height of D above ground at A 

= 175-16 + 4-25 = 179-41 ft 
Using Eq. (2. 59), 
™,a h\cot 2 a - cot 2 fi) + x 2 

COS <p = * 7T- — ^ 

^ 2hx cot a 

175-16 2 (cot 2 7°14'00 - cot 2 10° 15' 20") + 523-54 s 
2 x 175-16 x 523-54 x cot 7° 14' 00" 
= 0-85909 
<f> = 30° 47' 10" 

(b) Thus bearing of AD = 126° 03' 34' - 30° 47' 10" 

= 095° 16' 24" 

(c) Length of line AD y = h cot a 

= 175-16 cot 7° 14' 
= 1380-07 ft 



2.66 The broken base line problem 

Where a base line AD cannot be measured due to some obstacle 
the following system may be adopted, Fig. 2.40. 




Fig. 2.40 Broken base line 

Lengths x and z are measured. 

Angles a, j8 and 6 are measured at station E. 

To calculate BC = y : 

Method 1 



In triangle AEB EB = 
In triangle AEC EC = 



x sin EAB 

sin a 
(x + y) sin EAC 

sin (a + /S) 



SURVEYING TRIGONOMETRY 103 

Then S? = * sin ( a + ft) (2.62) 

EC (x + y) sin a 

Also in triangle EDB EB = Cv + z) sin EDB 

sin(0 + ft) 

in triangle EDC EC = z sinEDC 

sin# 

Then ER = fr + z > sin( ^ (2.63) 

EC zsin(0+ft) 

Equating Eqs (2.62) and (2.63) 

(y + z) sinfl _ x sin (a + ft) 
z sin (0 + ft) (x + y) sin a 

i.e. (x + yXy + z) = * z sin (a + ft) sin(fl + ft) 

sin a sin# 



Then yz + y( x + z) + xz 



1 - sin ( a + ft) sin ((9 + ft) 
sin a sin# 



= (2.64) 



This is a quadratic equation in y. Thus 

y = 



_ x + z + //*_+_z\ 2 _ ^[l _ sin(q + ft) sin(fl + ft) 1 
2 /J \ 2 J I sin a sin0 J 



±£+ /(* - Z ) 2 + r 7 Si "( a + ^ Sin (0 7ft) (265) 

2 V V 2 / sin a sin# 



Method 2 



Area of triangle 4 BE = Axfc = I^E.EB sin a (1) 

BCE = iyfc = IfiE. EC sin ft (2) 

CUE = ±zh = ICE. ED sin (9 (3) 

ADE = |(jc + y + z)h 

= ±AE . ED sin(a + ft + 0) (4) 

Dividing (1) by (2) x AE sin a 



y EC sin ft 



(5) 



Dividing (3) by (4), Z ~ - AB CE f Sin * _ (6) 

6 JV x + y + z AEsin(a + 8 + 6) 



104 SURVEYING PROBLEMS AND SOLUTIONS 

Multiplying (5) by (6), 

xz sin a sin0 



y(x + y + z) sin/3 sin(a + j8 + d) 

i.e. y* + y(x + z) - xz sin/3sin(a + /3 + fl) = 

sin a sin# 



Then y = -(*±±) + /fc±£V + xz sin P sin < a + Z 3 + ^> (2 
\ 2 / sj\ 2 / sina sinfl v ' 

Method 3 (Macaw's Method) 

In order to provide a logarithmic solution an auxiliary angle is 
used. 

From the quadratic equation previously formed, 



66) 



y 2 + y(x + z)- xz sin/3sin(a + /3 + fl) = (2 67) 

sina sin# 

i.e. {y + jfc, + z)P - J(, + zY + " Sin f in s ^°, + / + g) (2-68) 

Now let tan'M = 4«z si„ fl gin (a + p + (?) (2 69) 

(x + z) sin a sin 

Substituting this in Eq. (2.68), we get 

{ y + |(x + z)} 2 = I(x + z) 2 (l + tan 2 M) 

= i(jc + z) 2 sec 2 iW 

y + I(x + z) = |(x + z) secM 

y = I(x + z)(secM - 1) 

= (x + z) sec/Wkl - cosM) 

y = (x + z) secM sin 2 iiW (2.70) 

Example 2.13 The measurement of a base line AD is interrupted by 
an obstacle. To overcome this difficulty two points B and C were 
established on the line AD and observations made to them from a 
station E as follows: 

AEB = 20° 18' 20" 
BEC = 45° 19' 40" 
CED = 33° 24' 20" 

Length AB = 527-43 ft and CD = 685-29 ft. 

Calculate the length of the line AD. 

(R.I. C.S.) 



SURVEYING TRIGONOMETRY 105 

Here a = 20° 18' 20" \ a + p = 65° 38' 00" 

p = 45°19'40" a + p + d = 99°02'20" 

= 33° 24' 20" j p + d = 78° 44' 00" 

x = 527-43 | \{x + z) = 1(1212-72) = 606-36 

z = 685-29 J |(jc^z) = §(157-86) = 78-93 

By method 1 

y = -606-36 + As .93' , 527-43 x 685-29 sin 65° 38' sin 78° 44' 
V sin20°18'20"sin33°24'20" 

= -606-36 + V6230 + 1690057 
= -606-36 + 1302-415 
= 696-055 ft 

Then AD = 1212-72 + 696-055 = 1908 -775 ft 

By method 2 , 

y = -606-36 + yfa)6-36 2 + 527 ' 43 x 685 ' 29 sin 45°19'40" sin99°02'20" 
^ sin 20° 18' 20" sin 33° 24' 20" 

= -606-36 + V367672 + 1328 614 

= -606-36 + 1302-415 

= 696-055 ft 



By method 3 










By logs 


4 

X 

z 

sinp 






0-6020600 
2-7221648 
2-8358744 
9-851955 4 




sin (a + 


+ 0) 


9-994 573 1 




cosec a 






10-459 637 2 




cosec 






10-2591940 
6-725 458 9 




(x + z) 2 






6-1675210 




tanzM 






0-557 9379 




taniW 






0-278 968 9 _ M = 62° 15' 11" 
§-M = 31°07'36" 



106 



SURVEYING PROBLEMS AND SOLUTIONS 


secM 




0-332017 5 


sin~M 




9-713433 2 


sin ^M 

2 




9-713 433 2 


(*+ z) 




3-083760 5 
2-842644 4 




y 


= 696-055 



2.67 To find the relationship between angles in the horizontal and 
inclined planes (Fig. 2.41) 




Fig. 2.41 

Let (1) lines /4B, and B,C be inclined to the horizontal plane by a 
and /S respectively. 

(2) Horizontal angle ABC = 6 

(3) Angle in inclined plane AB^C = <f> 

(4) B,B = h 

Then AB = h cot a AB, = h coseca 

BC = h cot/8 B,C = fccosec/3 

In triangle ABC, 

4C 2 = AB 2 + BC 2 - 2 AB.BC cos (9 

= h 2 cot 2 a+ h 2 cot 2 fi - 2 A 2 cot a cot /8 cos 
Similarly in triangle AB,C, 

AC 2 = h 2 cosec 2 a + /z 2 cosec 2 /3 - 2fc 2 coseca cosec/S cos<£ 



SURVEYING TRIGONOMETRY 107 

Then 

h z cot 2 a + /i 2 cot 2 ft 

-2fc 2 cotacotft cos0 = /i 2 cosec 2 a + h 2 cosec 2 B 

- 2h z coseca cosecft cos<£ 

cos *0 = (cosec 2 a - cot 2 a) + (cosec 2 ft - cot 2 ft) + 2 cot a cot ft cosfl 

2 coseca cosecft 
cosec 2 a - cot 2 a = cosec 2 ft - cot 2 ft = 1 



as 



Then cos<£ = 2(1 + cot a cot ft cosfl) 

2 coseca cosecft 



or cos 



= sin a sin/8 + cos a cos ft cos# (2.71) 

6 = cos - sin a sin ft ,n y<y\ 

cos a cos ft 



Example 2.14 From a station 4 observations were made to stations 
B and C with a sextant and an abney level. 

With sextant - angle BAC = 84° 30' 
With abney level - angle of depression (AB) 8° 20' 
angle of elevation (AC) 10° 40' 
Calculate the horizontal angle BA^ C which would have been 
measured if a theodolite had been used 

(R.I.C.S./M) 
From equation (2.72), 

cos0 = cos - sin a sin ft 
cos a cos ft 

= cos 84° 30' - sin (-8° 20') sin 10° 40' 
cos (-8° 20') cos 10° 40' 

cos 84° 30' + sin8°20'sinl0°40' 
cos 8° 20' cos 10° 40' 

0-09585 + 0-14493 x 0-18509 
0-989 44 x 0-98272 

0-12268 
0-97234 

= 0-12617 

d = 82° 45' 10" 

Example 2. 15 A pipe-line is to be laid along a bend in a mine roadway 
ABC. If AB falls at a gradient of 1 in 2 in a direction 036° 27', whilst 
BC rises due South at 1 in 3-5, calculate the angle of bend in the pipe. 

(R.I.C.S.) 



108 



SURVEYING PROBLEMS AND SOLUTIONS 
3.5 




Fig. 2.42 



Plan 



From equation (2.71), 

cos0 = sin a sinjS + cos a cos/3 cos# 
where 

a = cot" 1 2 = 26° 33' 
£ = cot" 1 3-5 = 15° 57' 
6 = 036° 27' - 00° = 36° 27' 
.-. cos0 = sin 26° 33' sin 15° 57'+ cos 26° 33' cos 15° 57' cos 36° 27' 
cf> = 35° 26' 40" i.e. 35° 27' 

Exercises 2(b) 

8. Show that for small angles of slope the difference between hori- 
zontal and sloping lengths is h z /2l (where h is the difference of verti- 
cal height of the two ends of a line of sloping length I) 

If errors in chaining are not to exceed 1 part in 1000, what is the 
greatest slope that can be ignored ? 

[L.U/E Ans. 2° 34'] 

9. The height of an electricity pylon relative to two stations A and 
B (at the same level) is to be calculated from the data given below. 
Find the height from the two stations if at both stations the height of 
the theodolite axis is 5'— 0". 




SURVEYING TRIGONOMETRY 109 

Pylon Data: AB = 200 ft 

Horizontal angle at A = 75° 10' 

at B = 40° 20' 

Vertical angle at A = 43° 12' 

at B = 32° 13' 

A " ' x fl (R.I.C.S. Ans. mean height 139-8 ft) 

10. X, Y and Z are three points on a straight survey line such that 
XY = 56 ft and YZ = 80 ft. 

From X, a normal offset was measured to a point A and X/4 was 
found to be 42ft. From Y and Z respectively, a pair of oblique offsets 
were measured to a point B, and these distances were as follows: 
YB = 96 ft, ZB = 88 ft 

Calculate the distance 4fi, and check your answer by plotting to 
some suitable scale, and state the scale used. 

(E.M.E.U. Ans. 112-7 ft) 

11. From the top of a tower 120 ft high, the angle of depression of a 
point A is 15°, and of another point B is 11°. The bearings of A and 
B from the tower are 205° and 137° respectively. If A and B lie in a 
horizontal plane through the base of the tower, calculate the distance 
AB. 

(R.I.C.S. Ans. 612 ft) 

12 A, 6, C, D are four successive milestones on a straight horizon- 
tal road. 

From a point due W of A, the direction of B is 84°, and of D 
is 77°. The milestone C cannot be seen from 0, owing to trees. If 
the direction in which the road runs from A to D is 0, calculate 0, 
and the distance of from the road. 

(R.I.C.S. Ans. = 60°06' 50", OA = 3-8738 miles) 

13. At a point A, a man observes the elevation of the top of a tower 
B to be 42° 15'. He walks 200 yards up a uniform slope of elevation 
12° directly towards the tower, and then finds that the elevation of B 
has increased by 23° 09'. Calculate the height of B above the level 
of A. 

(R.I.C.S. Ans. 823-82 ft) 

14. At two points, 500 yards apart on a horizontal plane, observations 
of the bearing and elevation of an aeroplane are taken simultaneously. 
At one point the bearing is 041° and the elevation is 24°, and at an- 
other point the bearing is 032° and the elevation is 16°. Calculate 
the height of the aeroplane above the plane. 

(R.I.C.S. Ans. 1139ft) 



110 SURVEYING PROBLEMS AND SOLUTIONS 

15. Three survey stations X, Y and Z lie in one straight line on 
the same plane. A series of angles of elevation is taken to the top of 

a colliery chimney, which lies to one side of the line XYZ. The angles 
measured at X, Y and Z were: 

at X, 14° 02'; at Y, 26° 34'; at Z, 18° 26' 

The lengths XY and YZ are 400 ft and 240 ft respectively. 
Calculate the height of the chimney above station X. 

(E.M.E.U. Ans. 112-0 ft) 

16. The altitude of a mountain, observed at the end A of a base line 
AB of 2992-5 m, was 19° 42' and the horizontal angles at A and B 
were 127° 54' and 33° 09' respectively. 

Find the height of the mountain. 

(Ans. 1804 m) 

17. It is required to determine the distance between two inaccessible 
points A and B by observations from two stations C and D, 1000 m 
apart. The angular measurements give ACB = 47°, BCD = 58°, 

BDA = 49°; ADC = 59°. 

Calculate the distance AB 

(Ans. 2907 -4 m) 

18. An aeroplane is observed simultaneously from two points A and 
B at the same level, A being a distance (c) due north of B. From A 
the aeroplane is S^°E and from B N <f> E. 

Show that the height of the aeroplane is 

c tan a sin <f> 
sin(0 + <f>) 

and find its elevation from B. 

/L.U. Ans. B = tan-' sin<fttana \ 
V sintf / 

19. A straight base line ABCD is sited such that a portion of BC 
cannot be measured directly. If AB is 575-64 ft and CD is 728-56 ft 
and the angles measured from station to one side of ABCD are 

DOC = 56° 40' 30" 
COB = 40° 32' 00" 
BOA = 35° 56' 30" 

Calculate the length BC. (E.M.E.U. Ans. 259-32 ft) 

20. It is proposed to lay a line of pipes of large diameter along a 

roadway of which the gradient changes from a rise of 30° to a fall of 

10° coincident with a bend in the roadway from a bearing of N 22° W 

to N 25° E. 

Calculate the angle of bend in the pipe. 11M , n( on » N 

(Ans. ny J? ou ) 



SURVEYING TRIGONOMETRY 111 

21. At a point A at the bottom of a hill, the elevation of the top of 

a tower on the hill is 51° 18'. At a point B on the side of the hill, and 
in the same vertical plane as A and the tower, the elevation is 71°40'. 
AB makes an angle 20° with the horizontal and the distance AB = 52 
feet. Determine the height of the top of the tower above A. 

(L.U. Ans. 91-5 ft) 

22. Two points, A, B on a straight horizontal road are at a distance 
400 feet apart. A vertical flag-pole, 100 feet high, is at equal distances 
from A and B. The angle subtended by AB at the foot C of the pole 
(which is in the same horizontal plane as the road) is 80°. 

Find (i) the distance from the road to the foot of the pole, 
(ii) the angle subtended by AB at the top of the pole. 

(L.U. Ans. (i) 258-5 ft, (ii) 75° 28') 

Bibliography 

USILL, G.w. and HEARN, G., Practical Surveying (Technical Press). 
HADDOCK, M.H., The Basis of Mine Surveying (Chapman and Hall). 
LONEY, s.c, Plane Trigonometry (Cambridge University Press). 
blakey, J., Intermediate Pure Mathematics (Macmillan). 



3 CO-ORDINATES 



A point in a plane may be defined by two systems: 

(1) Polar co-ordinates. 

(2) Rectangular or Cartesian co-ordinates. 

3.1 Polar Co-ordinates 

This system involves angular and linear values, i.e. bearing and 
length, the former being plotted by protractor as an angle from the 
meridian. 



B (s,8) 




Fig. 3. 1 Polar co-ordinates 

A normal 6 inch protractor allows plotting to the nearest 1/4°; a 
cardboard protractor with parallel rule to 1/8°; whilst the special Book- 
ing protractor enables 01' to be plotted. 

The displacement of the point being plotted depends on the physi- 
cal length of the line on the plan, which in turn depends on the horiz- 
ontal projection of the ground length and the scale of the plotting. 




A s 

Fig. 3.2 Displacement due to angular error 
If a is the angular error, then the displacement 

112 



CO-ORDINATES 113 

BB X = s tana 
and as a is small 

BB, ~ sa 

If s = 300 ft and a = 01' 00", 

DD 300 x 60 x 12 . , 
fiS ' = 206 265 inches 

~ 1 inch 

i.e. 1 minute of arc subtends 1 inch in 100 yards, 

1 second of arc subtends 1 inch in 6000 yards, i.e. 3 x /a miles. 

Similarly, on the metric system, if s = 100 metres and a = 01 ' 00", 

♦i. dd 100x60 . 
then BB, = metres 

206 265 

= 0-0291 m, i.e. 29 mm 

Thus, 1 minute of arc subtends approximately 30 mm in 100m, 

1 second of arc subtends approximately 1 mm in 200 m 

or 1cm in 2 km. 

A point plotted on a plan may be assumed to be 0*01 in., (0*25mm), 

i.e. 0'Olin. in 1 yard (0*25 mm in 1 metre) — > 1 minute of arc, 

0*1 in. in 1 yard (25 mm in 1 metre) — * 10 minutes of arc. 

As this represents a possible plotting error on every line, it can be 
seen how the error may accumulate, particularly as each point is depend- 
ent on the preceding point. 

3.11 Plotting to scale 

The length of the plotted line is some definite fraction of the ground 
length, the 'scale' chosen depending on the purpose of the plan and the 
size of the area. 

Scales may be expressed in various ways: 

(1) As inches (in plan) per mile, e.g. 6 in. to 1 mile. 

(2) As feet, or chains, per inch, e.g. 10ft to 1 inch. 

(3) As a representative fraction 1 in n, i.e. 1/n, e.g. 1/2500. 



3.12 Conversion of the scales 

I insult ranracanfc ■.-— .- 

40 



40 inches to 1 mile — 1 inch represents — -rp: — feet 



lin. = 132 ft 
= 44 yd 
= 2 chn 



114 SURVEYING PROBLEMS AND SOLUTIONS 

1 inch to 132 ft - 1 inch represents 132 x 12 inches 

lin. = 1584 in. 
Thus the representative fraction is 1/1584. 



3.13 Scales in common use 

Ordinance Survey Maps and Plans: 

Large scale: 1/500, 1/1250, 1/2500. 

Medium scale: 6 in. to 1 mile (1/10 560), 2y 2 in. to 1 mile (1/25000). 

Small scale: 2,l, 1 / 2 , 1 / 4 in. to 1 mile; 1/625000, 1/1250000. 

Engineering and Construction Surveys: 

1/500, 1/2500, 10-50 ft to 1 inch, 1/4, V8, Vl6in. to 1ft. 
(See Appendix, p. 169) 

3.14 Plotting accuracy 

Considering 0*01 in. (0*25 mm) as the size of a plotted point, the 
following table shows the representative value at the typical scales. 



O.S. Scales 






suggested measurement 
precision limit 


1/500 


0-01 x 500 


= 5 in. 


3 in. (76 mm) 


1/1250 


0-01 x 1250 


= 12-5 in. 


1ft (0 -3 m) 


1/2500 


0-01 x 2500 


= 25-0 in. 


2 ft (0-6 m) 


1/10 560 


0*01 x 10 560 


= 105-6 in. 


5 ft (1-5 m) 


1/25000 


0*01 x 25000 


= 250-0 in. 


10 ft (3-0 m) 


Engineering Scales 






1 in. to 10 ft 


0-01 x 120 


= l-2in. 


lin. 


lin. to 50ft 


0-01 x 600 


= 6-0 in. 


6 in. 


lin. to lchn 


0-01 x 792 


= 7-92 in. 


6 in. or l / 2 link 


lin. to 2chn 


0-01 x 1584 


= 15-84 in. 


1 ft or 1 link. 



3.15 Incorrect scale problems 

If a scale of 1/2500 is used on a plan plotted to scale 1/1584 
what conversion factor is required to 

(a) the scaled lengths, 

(b) the area computed from the scaled length? 

(a) On the plan 1 in. = 1584 in. whereas the scaled value shows 
1 in. = 2500 in. 

All scaled values must be converted by a factor 1584/2500 
= 0-6336. 



CO-ORDINATES 



115 



(b) All the computed areas must be multiplied by (0*6336jf 
0-4014 



3.2 Bearings 

Four meridians may be used, Fig. 3.3: 

1. True or geographical north. 

2. Magnetic north. 

3. Grid north. 

4. Arbitrary north. 
GN. T . N . 

4 



M.N 




M.N. 



T.N. 



6.N. 



Convergence of 
the meridian 



Magnetic declination 




Fig. 3. 3 Meridians 



3.21 True north 



The meridian can only be obtained precisely by astronomical 
observation. The difference between true bearings at A and B is the 
convergence of the meridians to a point, i.e. the north pole. For small 
surveys the discrepancy is small and can be neglected but where 
necessary a correction may be computed and applied. 



3.22 Magnetic north 

There is no fixed point and thus the meridian is unstable and sub- 
jected to a number of variations (Fig. 3.4), viz.: 

(a) Secular variation — the annual change in the magnetic declina- 
tion or angle between magnetic and true north. At present the magnetic 
meridian in Britain is to the west of true north but moving towards it at 
the approximate rate of lOmin per annum. (Values of declination and 



116 



SURVEYING PROBLEMS AND SOLUTIONS 



the annual change are shown on certain O.S. sheets.) 

(b) Diurnal variation — a daily sinusoidal oscillation effect, with 
the mean value at approximately 10a.m. and 6-7 p.m., and maxima and 
minima at approximately 8a.m. and 1p.m. 

(c) Irregular variation — periodic magnetic fluctuations thought to 
be related to sun spots. 



Secular variation 




13001 



25* 20 15 10 5 5 10 15* 
West East 



Diurnal variation 



East 




0.00 24.00 Hours 
G.M.T. 



Fig. 3.4 Approximate secular and diurnal variations in magnetic 
declination in the London area 
(Abinger Observatory) 

3.23 Grid north (see section 3.7). 

O.S. sheets are based on a modified Transverse Mercator projec- 
tion which, within narrow limits, allows: 

(a) Constant bearings related to a parallel grid. 

(b) A scale factor for conversion of ground distance to grid dis- 
tance solely dependent on the easterly co-ordinates of the measurement 
site. (See page 39). 



3.24 Arbitrary north 

This may not be necessary for absolute reference and often the 
first leg of the traverse is assumed to be 0°00 '. 



CO-ORDINATES 



117 



Example 3.1 True north is 0°37'E of Grid North. 

Magnetic declination in June 1955 was 10°27' W. 
If the annual variation was 10' per annum towards North and the 
grid bearing of line AB 082°32' , what will be the magnetic bearing of 
line AB in January 1966? 

G.N. 
MM'66 
MK'55 »• 




Fig. 3.5 

Grid bearing 
Correction 

True bearing 

Mag declination 
June 1955 

Mag. bearing 
June 1955 

Variation for 
January 1966 
-lO^xlO' 

Mag. bearing January 1966 



082°32' 
-0°37 / 

081°55' 

10°27' 

092°22' 

-1°45' 
090°37' 



3.25 Types of bearing 

There are two types in general use: 

(a) Whole circle bearings (W.C.B.), which are measured clockwise 
from north or 0°-360°. 



118 



SURVEYING PROBLEMS AND SOLUTIONS 



(b) Quadrant bearings (Q.B.), which are angles measured to the 
east or west of the N/S meridian. 

For comparison of bearings, see Fig. 3.6. 

N 



Case (i) 



Case (ii) 




Case (Hi) 



Case (iv) 



W 270'- 



W 270- 




Fig. 3.6 Comparison of bearings 



CO-ORDINATES 119 

Case (i) Whole circle bearing in the first quadrant — 90° 

W.C.B. of AB = a° 

Q.B. of AB = N a?E 
Case (ii) 90°- 180° 

W.C.B. of AC = a° 2 

Q.B. of AC = S 0°E 

= S(180-a 2 )°E 

Case (Hi) 180°- 270° 

W.C.B. of AD = a° 3 

Q.B. of AD = S0°W 

= S(03-180)°W 

Case(iv) 270°- 360° 

W.C.B. of AE = a° 

4 

Q.B. of AE = N W 

= N (360 -a^W 

Example 3.2 

072° = N 72° E 

148° = S32°E i.e. 180-148 = 32° 

196° = S 16° W i.e. 196-180 = 16° 

330° = N30°W i.e. 360-330 = 30° 

N.B. Quadrant bearings are never from the E/W line, so that the pre- 
fix is always N or S. 

It is preferable to use whole circle bearings for most purposes, the 
only advantage of quadrant bearings being that they agree with the 
values required for trigonometrical functions 0-90° as given in many 
mathematical tables (see Chapter 2), e.g.: 

(Fig. 3.7a) sin 30° = 0-5 

cos 30° = 0-8660 
tan 30° = 0-5774 

(Fig. 3.7b) sin 150° = sin (180 -150) 
= sin 30° 

cos 150° = -cos (180 -150) 
= -cos 30° 

tan 150°= Sinl5 ° = +sin3 ° 
cos 150 -cos 30 

= -tan 30° 



120 



SURVEYING PROBLEMS AND SOLUTIONS 



(a) 



Cos 30° 




* E 



Bearing N E 
+ + 



(b) 



-Cos 30° 




Bearing S E 



*E 




210° 



-Cos 30* 



Bearing S W 




Bearing N W 
+ Cos 30° + ~ 



330" 



Fig. 3.7 



(Fig. 3.7c) sin 210° = - sin (210 - 180) 

= -sin 30° 

cps210° = -cos (210- 180) 

= -cos 30° 



CO-ORDINATES 



121 



tan 210° = 



sin 210 



cos 210 
= +tan30 c 



- sin 30 
-cos 30 



(Fig. 3.7d) sin 330° = - sin (360 - 330) 

= - sin 30° 

cos 330° = cos (360 -330) 

= + cos 30° 



tan 330° = 



sin 330 



- sin 30 



cos 330 + cos 30 
= -tan 30° 



3.26 Conversion of horizontal angles into bearings. (Fig. 3.8) 




Fig. 3.8 Conversion of horizontal angles into bearings 



Forward Bearing AB = a° 

Back Bearing BA = a ± 180° 

Forward Bearing BC = a ± 180 + 

If the sum exceeds 360° then 360 is subtracted, i.e. 

Bearing BC(j8) = a ± 180 + - 360 = a + Q ± 180 

This basic process may always be used but the following rules 
simplify the process. 

(1) To the forward bearing add the clockwise angle. 

(2) If the sum is less than 180° add 180°. 

If the sum is more than 180° subtract 180°. 
(In some cases the sum may be more than 540°, then subtract 
540°) 
N.B. If the angles measured are anticlockwise they must be sub- 
tracted. 



122 SURVEYING PROBLEMS AND SOLUTIONS 

Example 3.3 




Fig. 3.9 



bearing 


AB = 


030° 


+ angle 


ABC = 


210° 
240° 




- 


180° 


bearing 


BC = 


060° 


+ angle 


BCD = 


56° 
116° 




+ 


180° 


bearing 


CD = 


296° 


+ angle 


CDE = 


332° 
628° 




- 


540° 


bearing 


DE = 


088° 



N 30° E 



N60°E 



N64°W 



N88°E 



CO-ORDINATES 123 



Check bearing AB = 030° 

angles = 210° 

56° 

332° 

628° 
-«xl80°, i.e. -3x180° - 540° 



bearing DE = 088° 

The final bearing is checked by adding the bearing of the first line 
to the sum of the clockwise angles, and then subtracting some multiple 
of 180°. 

Example 3.4 The clockwise angles of a closed polygon are observed 
to be as follows: 

A 223°46' 

B 241°17' 

C 257°02' 

D 250°21' 

E 242° 19' 

F 225°15' 
If the true bearings of BC and CD are 123° 14' and 200° 16' 
respectively, and the magnetic bearing of EF is 333°2l', calculate 
the magnetic declination. 

(N.R.C.T.) 

From the size of the angles it may be initially assumed that these 
are external to the polygon and should sum to (2n + 4)90°", i.e. 

l(2x6) + 4}90 = 16x90 = 1440° 

223°46' 
241° 17' 
257°02' 
250° 21' 
242° 19' 
225°15' 

Check 1440°00' 
To obtain the bearings, 

Line BC bearing 123° 14' 

+ angle BCD 257°02' 

380° 16' 
- 180° 



124 



SURVEYING PROBLEMS AND SOLUTIONS 



bearing 


CD 


200° 16' 


(this checks with given 


+ angle 


CDE 


250°21' 
450°37' 


value) 




- 


180° 




bearing 


DE 


270°37' 




+ angle 


DEF 


242° 19' 






512°56' 






EF 


180° 




bearing 


332°56' 




+ angle 


EFA 


225° 15' 






558° 11' 






FA 


540° 




bearing 


018°11' 




+ angle 


FAB 


223°46' 






241°57' 






AB 


180° 




bearing 


061°57' 




+ angle 


ABC 


241°17' 






303° 14' 






BC 
EF 


180° 




bearing 


123° 14' 


Check 


Magnetic bearing 


333°21' 




True bearing 


EF 


332°56' 




Magnetic declination 


0°25' 


W 



3.27 Deflection angles (Fig. 3.10) 

In isolated cases, deflection angles are measured and here the 
normal notation will be taken as: 

Right angle deflection— positive. 
Left angle deflection— negative. 



Taking the Example 3.3, 

Bearing AB 
Deflection right 
Deflection left 
Deflection right 



030° 
+ 30° 
-124° 
+ 152° 



CO-ORDINATES 



125 



•152°(R) 




Fig. 3. 10 Deflection angles 



Bearing 


AB 


030° 
+ 30° 


Bearing 


BC 


060° 
+ 360° 

420° 
-124° 


Bearing 


CD 


296° 
+ 152° 

448° 
-360° 


Bearing 


DE 


088° 


Check 


AB 


030° 
+ 30° 
+ 152° 

+ 212° 
-124° 




DE 


088° 



126 SURVEYING PROBLEMS AND SOLUTIONS 

Exercises 3(a) 

1. Convert the following whole circle bearings into quadrant bearings: 

214°30' ; 027°15' ; 287°45' ; 093°30' ; 157°30'; 
311°45' ; 218°30' ; 078°45' ; 244°14' ; 278°04.' 
(Ans. S34°30' W; N 27°15' E; N 72°15' W; S 86°30' E; 

S 22°30' E; N 48°15' W; S 38°30' W; N 78°45' E; 

S64°14' W; N81°56' W ) 

2. Convert the following quadrant bearings into whole circle bearings: 

N 25°30' E; S 34°15' E; S 42°45' W; N 79°30' W; 
S 18°15' W; N 82°45' W; S 64°14' E; S 34°30' W. 

(Ans. 025°30' ; 145°45' ; 222°45' ; 280°30' ; 198°15' ; 
277°15' ; 115°46' ; 214°30') 

3. The following clockwise angles were measured in a closed tra- 
verse. What is the angular closing error? 

163°27'36"; 324°18'22"; 62°39' 27" ; 330° 19' 18" ; 
181°09'15"; 305°58'16"; 188°02'03"; 292°53'02"; 
131°12'50" (Ans. 09") 

4. Measurement of the interior anticlockwise angles of a closed tra- 
verse ABC D E have been made with a vernier theodolite reading to 
20 seconds of arc. Adjust the measurements and compute the bearings 
of the sides if the bearing of the line AB is N 43° 10' 20" E. 

Angle EAB 135°20'40" (R.I.C.S. Ans. AB N 43°10' 20" E 

ABC 60°21'20" BC S 17°10'52"E 

BCD 142°36'20" CD S 20°12'56" W 

CDE 89°51'40" DE N 69°38' 36" W 

DEA 111°50'40" EA N01°29'08" W) 

5. From the theodolite readings given below, determine the angles 
of a traverse ABCDE. Having obtained the angles, correct them to 
the nearest 10 seconds of arc and then determine the bearing of BC if 
the bearing of AB is 45° 20' 40". 



Back Station 


Theodolite 


Forward 


Readings 




Station 


Station 


Back Station 


Forward Station 


E 


A 


B 


0°00'00" 


264°49'40" 


A 


B 


C 


264°49'40" 


164° 29' 10" 


B 


C 


D 


164° 29' 10" 


43°58'30" 


C 


D 


E 


43°58' 30" 


314° 18 '20" 


D 


E 


A 


314°18' 20" 


179°59' 10" 



(R.I.C.S. Ans. 125°00' 20") 



CO-ORDINATES 



127 



3.3 Rectangular Co-ordinates 

A point may be fixed in a plane by linear values measured 
parallel to the normal xy axes. 

The x values are known as Departures or Eastings whilst the y 
values are known as Latitudes or Northings. 

The following sign convention is used: 



Direction 



East + x 

West -x 

North +y 

South _ y 



+ departure 

- departure 

+ latitude 

- latitude 



+ Easting (+E) 
-Easting (-E) 

+ Northing (+N) 
-Northing (-N) 



N.W. quadrant (270*-360*) N 
Latitude 



N.E. quadrant (0*-90') 



270* W-+ 




Departure (eastings) 
*90*E 



(-2,-4) 



S.W. quadrant (180'- 270*) 



(+4,-2) 



S.E. quadrant OC-ieO*) 



Fig. 3.11 Rectangular co-ordinates 



N.B. 



0°-90° 


-+ NE 


i.e. 


+ N +E 


or 


+ lat 


+ dep. 


90°- 180° 


— S E 


i.e. 


-N +E 




-lat 


+ dep. 


180°- 270° 


-» S W 


i.e. 


-N -E 




-lat 


-dep. 


270°- 360° 


-» NW 


i.e. 


+ N-E 




+ lat 


-dep. 



This gives a mathematical basis for the determination of a point 
with no need for graphical representation and is more satisfactory for 
the following reasons: 

(1) Each station can be plotted independently. 

(2) In plotting, the point is not dependent on any angular measur- 
ing device. 

(3) Distances and bearings between points can be computed. 



128 SURVEYING PROBLEMS AND SOLUTIONS 

Rectangular co-ordinates are sub-divided into: 

(1) Partial Co-ordinates, which relate to a line. 

(2) Total Co-ordinates, which relate to a point. 

3.31 Partial co-ordinates, AE, AN (Fig. 3.12) 

These relate one end of a line to the other end. 
They represent the distance travelled East (+)/West (-) and 
North (+)/South (-) for a single line or join between any two points. 




Fig. 3.12 Partial co-ordinates 
Given a line of bearing d and length s, 

Partial departure = AE i.e. difference in Eastings 
A E AB = s sin Q 

Partial latitude = AN i.e. difference in Northings 
AN^g = scosd 



(3.1) 



(3.2) 



N.B. always compute in bearings not angles and preferably quadrant 
bearings. 

3. 32 Total co-ordinates (Fig. 3. 13) 

These relate any point to the axes of the co-ordinate system used. 
The following notation is used: 

Total Easting of A = E A 
" Northing of A = N^ 

Total Easting of B = E A + A E AB 
« Northing of B = N A + AN AB 

Total Easting of C = E B + A E BC = E A + A E AB + A E BC 
" Northing of C = N B + &N BC = N^ + AN^ + AN BC 



CO-ORDINATES 



129 




Fig. 3.13 Total co-ordinates 

Thus in general terms 

Total Easting of any point = E^+SAE (3.3) 

= Total easting of the first 
point + the sum of the 
partial eastings up to that 
point. 

Total Northing of any point = N^ + 2 AN (3.4) 

= Total northing of the first 
point + the sum of the 
partial northings up to 
that point. 

N.B. If a traverse is closed polygonally then 

SAE = (3.5) 

SAN = (3.6) 

i.e. the sum of the partial co-ordinates should equal zero. 

Example 3.5 

Given: (Fig. 3.14) AB 045° 100 m 

BC 120° 150 m 
CD 210° 100 m 
Total co-ordinates of A E 50 m N 40 m 

Line AB 045° = N 45° E 100 m 
Partial departure AE^ = 100 sin 45° = 100 x 0-707 = + 70*7 m 

Total departure (E^) A = + 50*0 m 
Total departure (E fl ) B= + 120-7 m 



130 



SURVEYING PROBLEMS AND SOLUTIONS 




-50* 



(200-6.-50-9) 



Fig. 3.14 



Partial latitude A N^ = 100 cos 45°= 100x0*707 = + 70-7 m 

Total latitude (N^) A = + 40*0 m 
Total latitude (N 5 ) B= + 110*7 m 

Line BC 120° = S 60° E 150 m 
Partial departure AE BC = 150 sin 60° = 150 x 0*866 = + 129*9 m 

Total departure (E B ) B = + 120*7 m 
Total departure (E C )C = + 250*6 m 

Partial latitude AN^ C = 150 cos 60° = 150 x 0*5 = - 75*0 m 

Total latitude (N B )fi = +110*7 m 
Total latitude (N c ) C = + 35*7 m 

Line CD 210°= S30°W 100 m 
Partial departure AE CD = 100 sin 30° = 100 x 0*5 = - 50*0 m 

Total departure (E^) C = + 250*6 m 

Total departure (E^) D = + 200*6 m 

Partial latitude AN CD = 100 cos 30° = 100 x 0*866 = - 86*6m 

Total latitude (N c ) C = + 35*7 m 
Total latitude (N^) D = - 50*9 m 
Check E D = E A + &E AB + AE BC + AE CZ? 

= 50*0 + 70*7 + 129*9 - 50*0 = + 200*6 m 



CO-ORDINATES 131 

N^= K A + Mi AB + AN SC + ANqd 

= 40-0 + 70-7 - 75-0 - 86'6 = - 50-9 m 

Exercises 3(b) (Plotting) 

6. Plot the following traverse to a scale of 1 in = 100 links, and 
thereafter obtain the length and bearing of the line AB and the area 
in square yards of the enclosed figure. 

N 21° W 120 links from A 

N 28° E 100 links 

N60°E 117 links 

N 32° E 105 links 

S 15° E 200 links 

S 40° W 75 links to B 

(Ans. From scaling N62°45'E 340 links; approx. area 3906 sq yd,) 

7. The following table shows angles and distances measured in a 
theodolite traverse from a line AB bearing due South and of horizontal 
length 110 ft. 

Angle Angle value Inclination Inclined distance (ft) 

ABC 192°00' +15° BC 150 

BCD 92°15' 0° CD 200 

CDE 93°30' -13° DE 230 

DEF 170°30' 0° EF 150 

Compute the whole circle bearing of each line, plot the survey to 
a scale of 1 in. = 100 ft and measure the horizontal length and bear- 
ing of the closing line. 

(M.Q.B./M. Ans. 260 ft; 076°30') 

8. The following notes reter to an underground traverse made from 
the mouth, A, of a surface drift. 

Line Bearing Distance (links) 

AB 038° 325 dipping at 1 in 2-4 

BC 111° 208 level 

CD 006° 363 level 

DE 308° 234 rising at 1 in 3-2 

Plot the survey to a scale of 1 chain to 1 inch. 

Taking A as the origin, measure from your plan, the co-ordinates 
of E. 

What is the difference in level between A and E to the nearest 
foot? 

(M.Q.B./UM Ans. E, E 233 links N 688 links; diff. in level AE 
78 ft) 



132 SURVEYING PROBLEMS AND SOLUTIONS 

9. Plot the following notes of an underground traverse to a scale of 

lin ** 100 ft. 



Line 


Bearing 


Distance 


AB 


N 28° W 


354 ft dipping at 1 in 7 


BC 


N 83° W 


133 ft level 


CD 


S 83° W 


253ft level 


DE 


N 8°E 


219 ft rising at 1 in 4 


EF 


S 89° E 


100 ft level 



Points A,B,C and D are in workings of a lower seam and points 
£ and F are in the upper seam, DE being a cross measure drift be- 
tween the two seams. 

It is proposed to drive a drift from A to F. 

Find the bearing, length, and gradient of this drift. 

(M.Q.B./UM Ans. N 40° W; 655 ft; +1 in 212) 

10. The co-ordinates in feet, relative to a common point of origin 
A, are as follows: 

Departure Latitude 

A 

B 275 E 237 N 

C 552 E 230 N 

D 360 E 174 S 

Plot the figure A BCD to a scale of 1 inch to 100 ft and from the 
co-ordinates calculate the bearing and distance of the line AC. 

(M.Q.B./UM Ans. N 67°24' E; 598ft) 

11. An area in the form of a triangle ABC has been defined by the 
co-ordinates of the points A B and C in relation to the origin 0, as 
follows: 



A 


South 


2460 ft 


East 3410 ft 


B 


North 


2280 ft 


East 4600 ft 


C 


North 


1210 ft 


East 1210 ft 



Plot the positions of the points to a scale of 1 in. to 1000 ft, and 
find the area, in acres, enclosed by the lines joining AB, BC and 
CA. 

(M.Q.B./M Ans. 169-826 acres) 

12. There is reason to suspect a gross angular error in a five-legged 
closed traverse in which the recorded information was as follows: 

Interior angles: A 110°; B 150°; C 70°; D 110°; E 110° 

Sides: AB 180 ft; BC 420 ft; CD 350 ft; DE 410 ft; 

EA 245 ft 



CO-ORDINATES 



133 



Plot the traverse to a scale of 100 ft to lin. and locate the gross 
angular error*, stating its amount. 

(L.U./E) 

13. A rough compass traverse of a closed figure led to the following 
field record: 



Line 


Length 


Bearing 


AB 


422 


57° 


EC 


405 


316° 


CD 


348 


284° 


DE 


489 


207° 


EA 


514 


109° 



Plot the figure (scale lin = 50ft) and adjust it to close using a 
graphical method. Letter your plan and add a north point (magnetic 
declination 10° W). 

(L.U./E) 

3.4 Computation Processes 

As tables of trigonometrical functions are generally tabulated only 
in terms of angles 0°-90°, it is convenient to convert the whole circle 
bearings into reduced or quadrant bearings. 

The signs of the partial co-ordinates are then related to the sym- 
bols of the quadrant bearings, Fig. 3.15. 

E + 1 r^ N 

> Departures 

W - J 



::) 



Latitudes 



Alternatively, the whole circle 
bearings are used and the sign of 
the value of the partial co-ordinate 
is derived from the sign of the 
trigonometrical function. 



W 





+ 






N W 

+ - 




N E 
+ + 










+ 


sw 




SE 

- + 





— 

s 



Fig. 3. 15 



The process may be either : 
(i) by logarithms or 
(ii) by machine (using natural trigonometrical functions). 



See Chapter 6 on location of errors. 



134 SURVEYING PROBLEMS AND SOLUTIONS 

3.41 Computation by logarithms 

Let AB = 243° 27' 423-62 m (.4 2063-16 m E 5138 -42 m N) 
(243° 27' = S63°27' W) 
Logs E A 2063-16 m 

partial departure (AE) 2-578 579 

sin bearing AB 1-951 602 \ 
distance 2-626977 1\ 
cos bearing 1 -650 287 ) 
partial latitude (&N AB ) 2-277 264 

N.B. The log distance is written down once only, being added to the 
log sin bearing above and the log cos bearing below, to give the partial 
departure and latitude respectively. 

It is often considered good computing practice to separate the log 
figures for convenience of adding though the use of squared paper 
would obviate this. 

3.42 Computation by machine 



t^AB 


-378-95 


E B 


1684-21 m 


"a 


5138-42 


AN AB 


-189-35 


N B 


4949-07 m 







E A 


2063-16 


partial departure 


AR._ 


AB.o 


-378-95 






E B 


1684-21 m 


sin bearing 


0-894 545 J 
23-62 J \ 
0-446 979 1 




distance A 






cos bearing 


"a 


5138-42 


partial latitude 


AN > 


W AB 


189-35 


^MB * 








"b 


4949-07 in 



Using a normal digital machine, the distance (being common) is 
set once in the machine and then separately multiplied by the appropri- 
ate trigonometrical function. 

In the case of the twin-banked Brunsviga, the processes are simul- 
taneous. 

N.B. For both natural and logarithmic trigonometrical functions the 
following tables are recommended: 

Degrees only 4 figure tables 

Degrees and minutes 5 figure tables 

Degrees, minutes and seconds 6 figure tables 

Degrees, minutes, seconds and decimals of seconds 7 figure tables 



CO-ORDINATES 



135 



3.43 Tabulation process (Fig. 3.16) 

Nottingham Regional College of Technology 
Traverse computation sheet N* 1 Traverse A to Z 

Compiled by _^±y_- (Red) Computed by .^/i.jGreen) 

Date _?j_r„/ii.7 : . Checked by _!??•„. 

4E = length x sin B'g 4N= length x cos B'g 



Line 


Bearing 


Length 


sin/cos S£r 


AE 


AH 


E 


N 


Ftoint 


AB 


243* 27'00" 
S63*27'W 


423-62 


s 0-894 545 
C0446 979 


-378 95 


-189-35 


2063-16 


5138-42 


A 












1684-21 


494907 


B 


BC 


042* 32' 00" 
N42*32' E 


221-38 


SO676 019 
cO-736 884 


H49-66 


+163-13 










1833 87 


5112-20 


C 








Check £ 


-229-29 
2063-16 


-26-22 
5138-42 










1833-87 


5112-20 





Fig. 3. 16 Tabulated computation 

Example 3.6 Calculate the total co-ordinates, in feet, of a point B 

if the bearing of AB is 119° 45' and the distance is 850 links on a 

slope of 15° from the horizontal. 

The co-ordinates of A relative to a local origin are N 5356*7 ft 

E 264-5 ft. 

(M.Q.B./UM) 

To find horizontal length (Fig. 3.17) 




Fig. 3. 17 

AB, = AB cos 15° links 

but 100 links = 66 ft; therefore to convert links to feet the length must 
be multiplied by K = 0*66, i.e. 

AB, = K.AB cos 15° 

= 0-66 x 850 x cos 15° 
By logs, 0-66 1-81954 

850 2-92942 

cos 15° T-984 94 
AB, 2-73390 



136 



SURVEYING PROBLEMS AND SOLUTIONS 



To find partial co-ordinates of line AB (Fig. 3.18) 




Fig. 3. 18 
119° 45' = S60°15' E 



By logs, 



partial departure 2*672 52 



A£ 



'AB 



sin 60° 15' 

AB, 

cos 60° 15' 



1-938 62 

2-73390 (see above) 

1-695 67 



partial latitude 2-429 57 



AN 



+ 264-5 
+ 470-4 



E B +734-9 



5356-7 
-268-9 



'AB 

N D 5087-8 



Co-ordinates of B, N 5087-8 ft E 734-9 ft 

3.44 To obtain the bearing and distance between two points given 
their co-ordinates (Fig. 3. 19) 

Let the co-ordinates of A and B be E^N^ and E s N fl respectively. 



F _ F 
Then tan bearing (0) = —2 ± 

N - N 

™B n A 



AE 



•AB 



AN 



AB 



N.B. For convenience this is frequently written : 
Bearing AB = tan-'AE/AN 
(the sign of the differences will indicate the quadrant bearing). 
Length AB = >/(AE* + AN* ) 



(3.7) 
(3.8) 

(3.9) 
(3.10) 



CO-ORDINATES 



137 



N.B. This is not a very good solution for computation purposes and 
the trigonometrical solution below is preferred. 



or 



AB = 



AB = 



cos bearing (0) 
AN^g sec# 



AE 



AB 



(3.11) 
(3.12) 
(3.13) 



sin bearing (0) 

= AE^cosectf (3.14) 

If both of these determinations are used, their agreement provides 
a check on the determination of 0, but no check on the subtraction of 
the Eastings or Northings. 











AZab 






N„ 








1 'B 


*"ab 


, 


^a . 


^/* 




>B 


N/, 


• 












A 















E 


A 




E 


■B 



Fig. 3.19 To find the length and bearing between two points 



Example 3.7 



A 
B 



E 
632-16 m 
925-48 m 



N 
949-88 m 
421-74 m 



AE + 293-32 m AN - 528-14 



Bearing AB = tan"' + 293 ' 32 (E) 
-528-14 (S) 



138 



SURVEYING PROBLEMS AND SOLUTIONS 



By logs, 



293-32 2-467 34 

528-14 2-72275 

tan (0) 1-744 59 — 

Length AB = AN sec 6 

= 528-14 sec 29° 03' 



» S 29° 03' E 

i.e. 150^57' 
or A E cosec 6 

293-32 cosec 29° 03' 



By logs, 



or 



528-14 
sec 29° 03' 

293-32 
cosec 29° 03' 



2-72275 
0-058 39 
2-781 14 

2-46734 
0-31375 
2-78109 



-* 604-14 m 



-> 604-07 m 



The first solution is better as AN > AE, but a more compatible 
solution is obtained if the bearing is more accurately determined, using 
7 figure logs, 



or 



AE 


2-4673417 


AN 


2-7227491 


tan ((9) 


9-7445926 


Bearing (0) = 


29°02'50" 


AN 


2-7227491 


sec# 


10-058 3786 




2-7811277 


AE 


2-467 3417 


cosec 


10-3137836 



- 604-13 m 



2-7811253 



604- 12 m 



Example 3.8 The following horizontal angle readings were recorded 
during a counter-clockwise traverse ABCD. If the line AD is taken 
as an arbitrary meridian, find the quadrantal bearings of the remaining 
lines. 

Find also the latitudes and departures of the line CD whose 
length is 893-6 m. 



COORDINATES 



139 



Station at 


Sight Vernier A Vernier B 


A 


D 241° 36' 20" 061° 36' 40" 


A 


B 038° 54' 00" 218° 53' 40" 


B 


A 329° 28' 00" 149° 28' 20" 


B 


C 028° 29' 00" 208° 29' 00" 


C 


B 106° 58' 20" 286° 58' 40" 


C 


D 224° 20' 20" 044° 20' 20" 


D 


C 026° 58' 00" 206° 58' 40" 


D 


A 053° 18' 40" 233° 18' 00" 


Ans. Mean values 




A D 


Angle 
241° 36' 30" 


B 


038° 53' 50" DAB 157° 17' 20" 


B A 


329° 28' 10" 


C 


028° 29' 00" ABC 59°00'50" 


C B 


106° 58' 30" 


D 


224° 20' 20" BCD 117° 21' 50" 


D C 


026° 58' 20" 


A 


053° 18' 20" CDA 26°20'00" 


360° 00 '00" 


Bearing AD = 0°00' 


Angle DAB = 157° 17' 20" 


Bearing AB = 157° 17' 20" (S 22° 42' 40" E) 


Angle ABC = 59° 00' 50" 




216° 18' 10" 




-180° 


Bearing BC 036° 18' 10" (N 36° 18' 10" E) 


Angle BCD 117° 21' 50" 




153° 40' 00" 




+ 180° 


Bearing CD 333°40'00 y/ (N 26°20'00" W) 


Angle CD/ 


I 26° 20' 00" 


360° 00' 00" 


Co-ordinates CD 893-6 m (N26°20'W) 


AE = 


893-6 sin 26° 20' = -396-39 m 


AN = 


893-6 cos 26° 20' = +800-87m 



140 SURVEYING PROBLEMS AND SOLUTIONS 

Ans. AB = S 22° 42' 40" E 

BC = N36°18'10"E 
CD = N26°20'00" E 
Co-ordinates of line CD AE = -396 -4m AN = +800*9 m 

Example 3.9 In order to continue a base line AC to G, beyond a 
building which obstructed the sight, it was necessary to make a trav- 
erse round the building as follows, the angles being treated as deflec- 
tion angles when traversing in the direction ABCDEFG. 

ACD = 92° 24' to the left 

CDE = 90° 21' to the right CD = 56-2 ft 

DEF = 89° 43' to the right DE = 123-5 ft 

Calculate EF for F to be on AC produced and find EFG and CF. 

(L.U.) 

IG 

i 

I 

89°43' 

£ ^_J $__».-- 

r 
i 
i 
i 

I 
i 
i 
i 
i 
I 
I 
i 
X. 

1- 
i 
i 
I 
I 
I 
I 
I 
I 
i 
I 
I 
I 



90°21y 



92°24', 



1 -o- 



Fig. 3.20 
Assuming the bearing AC = 0°00' 



CO-ORDINATES 141 

i.e. Bearing AC = 360° 00' 

-angle ACD (left) 92° 24' 

Bearing CD 267° 36' i.e. S 87° 36' W 

+ angle CDE (right) 90 o 2l' 

Bearing DE 357° 57' i.e. N 02° 03' W 

+ angle DBF (right) 89° 43' 

447° 40' 
-360° 00' 



Bearing EF 087° 40' i.e. N 87° 40' E 

Thus, to obtain the bearing of FG = bearing AC, 
the deflection angle EFG = 87° 40' left. 
Check on deflection angles 

+ 
90° 21' 92° 24' 
89° 43' 87° 40' 



180° 04' 180° 04' 



To obtain the co-ordinates of E 



Line 


Distance (ft) 


Bearing 


sin Bearing 


cos Bearing 


Ae 


An 


AC 




o°oo' 










CD 


56-2 


S 87°36'w 


0-999 12 


0-04188 


-56-15 


- 2«35 


DE 


123-5 


N02°03'w 


0-035 77 


0-999 36 


- 4-42 


+ 123*42 



+ 123.42 
- 2.35 



E F - 60.57 N^ +121-07 

Thus F must be +60*57 ft east of E. The line EF has a bearing 
087° 40' 

Length EF = AE ^ = 60 ' 57 , 

sin bearing sin 87° 40' 

= 60-62 ft 

To find the co-ordinates of F, 

&N EF = 60-62 cos 87° 40' = + 2-47 

N £ = +121-07 

N/r = +123-54 



142 SURVEYING PROBLEMS AND SOLUTIONS 

F is 123*54 ft above C on the bearing due N. 
.-. CF = 123-54 ft. 



Ans. EF = 60-6 ft 

EFG = 87° 40' deflection left 
CF = 123-5 ft 

Example 3.10, The co-ordinates (metres) of the base line stations A 
and B are 

A 26 543-36 E 35432-31 N 

B 26895-48 E 35983-37 N 

The following clockwise angles were measured as part of a closed 
traverse: ABCDEA 

ABC 183° 21' 

BCD 86° 45' 

CDE 329° 17' 

DEA 354° 36' 

EAB 306° 06' 

Determine the adjusted quadrant bearings of each of the lines rela- 
tive to the meridian on which the co-ordinates were based. 

A 26 543-36 m 35 432-31 m 

B 26895-48 m 35 983-37 m 

AE 352-12 m AN 551-06 m 

tan bearing AB = 352-12 
551-06 

bearing AB = 032° 34' 

corr. Corrected Angle 

2 Angles 183° 21' -01' 183° 20' 

86° 45' -01' 86° 44' 

329° 17' -01' 329° 16' 

354° 36' -01' 354° 35' 

306° 06' -01' 306° 05' 

1260° 05' 1260° 00' 



£ Angles should equal (2n + 4) 90 

i.e. (2x5 + 4)90 = 1260< 
.*. error is 05' distributed as 01' per angle. 



CO-ORDINATES 143 



Calculation of bearings 






Bearing AB 


032° 34' - 


> N 32° 34' F 




Angle ABC 


183° 20' 
215° 54' 






- 180° 




Bearing BC 
Angle BCD 


035° 54' - 


> N ^5° 54' F 


86° 44' 






122° 38' 




+ 180° 




Bearing CD 


302° 38' 


* N57°22'W 


Angle CDE 


329° 16' 
631° 54' 
-540° 




Bearing DE 


091° 54' - 


i ^ ftR°nfi' f 




Angle DEF 


354° 35' 
446° 29' 
-180° 




Bearing EA 


266° 29' - 


» S 86°29'W 


Angle EAB 


306° 05' 
572° 34' 






-540° 




Bearing AB 


032° 34' 


Check 



Example 3.11 A disused colliery shaft C, situated in a flooded area, 
is surrounded by a circular wall and observations are taken from two 
points A and 6 of which the co-ordinates, in feet, relative to a local 
origin, are as follows: 

Station Eastings Northings 

A 3608-1 915-1 

B 957-6 1808-8 

C is approximately N.W. of A. 

Angles measured at A to the tangential points 1 and 2 of the walls 
are BAC, = 25° 55' and BAC Z = 26° 35'. 

Angles measured at B to the tangential points 3 and 4 of the wall 
are C 3 BA = 40° 29' and C A BA = 39° 31'. 

Determine the co-ordinates of the centre of the shaft in feet rela- 
tive to the origin, to one place of decimals and calculate the diameter 



144 SURVEYING PROBLEMS AND SOLUTIONS 

of the circle formed by the outside of the wall. 



A 
B 



AE 



'AB 



Fig. 3.21 
E 
3608-1 
957-6 
_ 2650-5 



N 
915-1 
1808-8 
AN„ D + 893-7 



VU3 



In Fig. 3.21, 



Bearing of AB = tan 



-iAE _ 



AN 



= tan" 



-2650-5 
+ 893-7 



(M.Q.B./S) 




= N71°22'W i.e. 288° 38' 

Length AB = AN sec bearing or AE cosec bearing 

= 893-7 sec71° 22' or 2650-5 cosec 71° 22' 

= 2797>1 2797-1 

In triangle ABC 



Angle A = I {25° 55'+ 26° 35'} 
Angle fi = I { 40° 29' +39° 31'} 
Angle C = 180° - (26° 15' + 40° 00') 



= 26° 15' 

= 40° 00' 

= 113° 45' 
180° 00' 



CO-ORDINATES 145 

By the sine rule, 

BC = AB sin A cosec C 

= 2797-1 sin 26° 15' cosec 113° 45' = 1351-6 ft 

AC = BC sin B cosec A 

= 1351-6 sin 40° 00' cosec 26° 15' = 1964-3ft 

Bearing AB 288° 38' 

+ Angle BAC 
Bearing AC 

Bearing BA 

- Angle CBA 

Bearing BC 

To find co-ordinates of C 

Line BC 068° 38' i.e. N68°38'E 1351-6 ft 



26° 


15' 


314° 


53' 


108° 
40° 


38' 
00' 


068° 


38' 



AE^ = 1351-6 sin 68° 38' 


= +1258-7 


E c = E s + ^E BC 




= 957-6 + 1258-7 


= +2216-3 


AN 5C = 1351-6 cos 68° 38' 


= + 492-4 


N c = N B + AN SC 




= 1808-8 + 492-4 


= +2301-2 



Check 

Line AC 314° 53' i.e. N45°07'W 1964-3 ft 

AE^ = 1964-3 sin 45° 07' = -1391-8 

E C = E A + & E AC 

= 3608-1 - 1391-8 = 2216-3 

AN^ C = 1964-3 cos 45° 07' = +1386-1 

N c = N A + AK AC 

= 915-1 + 1386-1 = 2301-2 

To find the diameter of the wall . 
Referring to Fig. 3. 21 

a = |(40 o 29'-39°31') = 0°29' 

R = BCsinO°29' cz BC x 0°29'(rad) 

= 1351-6 x 29 x 60 = 11.40 ft 
206265 



146 



SURVEYING PROBLEMS AND SOLUTIONS 



Check 



p = 1(26° 35' -25° 55') = 0°20' 

R = AC x 0°20'(rad) 

1964-3 x 20 x 60 = n^ft 

206 265 



Diameter of wall 



22-8 ft 



3.5 To Find the Co-ordinates of the Intersection of Two Lines 
3.51 Given their bearings from two known co-ordinate stations 

As an alternative to solving the triangle and then computing the 
co-ordinates the following process may be applied: 

Given A (E A N A ) 

B (E 5 N fl ) 
bearings a and jQ 



C (E C N C ) 




Fig. 3. 22 



From Fig. 3.22, 



(3.15) 
(3.16) 



E c = E A + (N^-N^tana 

= E A + AN^ tana 

= E B + (N c -N B )tan/8 

= E B + AN sc tan0 

.-. N c (tana -tanjS) = E B - E A + N^ tana - N B tan0 

Then the total northing of C 

N = E g - E^ + H A tana - N B tanft (3 1?) 

c ~ tana - tanjS 

To obtain the partial co-ordinates from equation (3.17) 



CO-ORDINATES 147 



Partial Northing Mi AC = N c - N A 

1,C N - N = Eg ~ Ea + Na tan a ~ Ng ta " ^ - N^ 
c A tana - tan/8 

E 5 - E^ + N^tan a - N^tanft - N^ tana + N A tanft 

tana - tanft 

_ (Es-E.0 -(N B -K4) tanft 



tan a - tan ft 

-/IS 

tana - tan/3 



Then AN^ = A E^, - A IWa. (3 lg) 



Then W BC = "T f. (3-19) 



Similarly, from equation (3.17), 

AU BC = N^-N* = Eg-E^ + N^tana-Nstanft 

tana - tanft 
& E AB - &N AB tan a 
tana - tan/8 

The following alternative process may be used: 

N c = N4 +(E c -E x )cota = N^ + AE^ cot a (3.20) 

= N B + (E c -E 5 )cotft = N B + AE 5C cotft (3.21) 

As before, the total and partial co-ordinates are given as: 

E = Nb ~ Na + Ea cot a ~ E B cot ft n 2? . 

C cot a - cot ft ^ ' 

and AE^ m *B ~ AE^cotft 



(3.23) 
(3.24) 



cot a - cot ft 
A E = AN4 g - AE^ cot a 
cot a - cot ft 

N.B. Theoretically, if Scot > Stan, then it is preferable to use the 
cot values, though in practice only one form would be used. 

Example 3.12 Let the co-ordinates be A = E4, N6 8 = E13, N4 
the bearings be a = 060° ft = 330° 

tana = 1-7321 cot a = 0-5774 

tan ft = -0-5774 cot ft = -1-7321 

Using the tan values ; 
from equation (3.17) 

N = (13 - 4) + (6 x 1-732 1) + (4 x 0-577 4) = g . 
from equation (3. 15) 



c = _ - - -• ' - = y-397 

1-7321 + 0-5774 



E c = 4 + (9-397-6) x 1*7321 = 9-884 



148 SURVEYING PROBLEMS AND SOLUTIONS 

or equation (3. 16) 

E c = 13 + (9-397 -4) x -0-5774= 9-884 
Using the cot values, 
from equation (3.22), 

E„ - 4 - 6 + 4 x 0-5774+ 13 x 1-7 321 Q .oo d 

C 0-577 4+1-7321 = ^ 

From equation (3.20), 

N c = 6 + (9-884 - 4) x 0-577 4 = 9-397 
or equation (3.21) 

N c = 4 + (9-884-13) x -1-7321 = 9-397 

If the formulae using partial values are employed the individual 
equation computation becomes simplified. 

Using the previous values (Ex. 3.5), 
from equation (3.18) 

W Ae . (13 -?-<1-?JL: 0,S774 - + 3-397 
AC 1-7321 + 0-577 4 

Then N c = N^ + AN^ 

= 6 + 3-397 = 9-397 

When this value is known, equation (3.15) may be used as before. 

From equation (3.23); 

(4-6) - (13-4) x -1-7321 



AEUc = 
E 



0-577 4 + 1-7321 

c = e a + ^AC 

= 4 + 5-884 = 9-884 



= +5-884 



When this value is known, equation (3.20) may be used as before. 
The above process is prefered and this can now be given in a 
tabulated form. 



Example 3.13 




(1) AN 



_ &E AB - AN^tan/8 

'AC — — — — — — — — ^— — — — I— 

tan a - tan /S 

(2) AE^ = tOA AC tana 

(3) AN 5e = AE^ B - b& A B tana 

tana - tan^S 

(4) AE BC = AN flC tan/3 



CO-ORDINATES 



149 



Igloo 



Oriented diagram 




E base 



Stations 


E 


Bearings 


N 


_E_B_ase_ ^AX 


+ 13 486-85 m 


a 278° 13' 57" 


+ 10 327-36 m 


_Iglpo__(B) 


+ 12 759-21 m 


182° 27' 44" 


+ 13 142-72 m 






tana -6-911745 2 




&AB 


-727-64 


tanjS + 0-0430004 
tan a - tan 


+ 2815-36 


AN^tan/8 


+ 121-06 








-848-70 


-J- -6-9547456" 
* x tana — 




&&AC 


-843-44 


+ 122-03 


W Base E c 


12 643-41 m 




10 449-39 m 






Check 




& E AB 


-727-64 






Afy^tana 


-19459-05 


tan a - tan /8 










+ 18 731-41 


+ -6-9547456" 
« — x tan/3 — ' 






& BC 


115-81 


2693-33 


W Base E c 


+ 12643-40 m 






10 449-39 m 



AN 



l AB 






AN, 



l BC 



3.52 Given the length and bearing of a line AB and all the angles 
A,B and C, Fig 3. 23 

Given: (a) Length and Bearing of AB, (b) Angles A,B and C. 
E c - E^ = b cos(A +6) 

= b(cosA cos 6 - sin A sin#) 

c sin B cos A cos^ - c sinB sin 4 sin 5 



but E« - E\d = c cos Q 



sinC 
j4B sin bearing^ 



SURVEYING PROBLEMS AND SOLUTIONS 
C 




Fig. 3. 23 



N, 



and 



Then 

E,7 — E. = 



. - N4 = c sintf = AB cos bearing^ 
C = 180 - 04 + B) 
sinC = sin A cosB + cos A sin 6 

(E B - E4) sinB cos 4 - (N B - N^) sing sin ,4 , 

sin A cos B + cos A sin S 
E4 sin A cos g + E^ cos A sin B + Eg sin B cos 4 - E^ sin g cos 4 - N B sin B sinA + Ha sin g sin A 



sin 4 cos g + cos A sin B 



E A cotB + E B cot 4 + (N.4 - N B ) 

cot A + cotB 
E.4 cotg + E fl cot /I - AN^s 



Similarly, 



cot 4 + cotB 



N^ = 



N,4 cotB + N B cot ,4 + &E AB 



(3.25) 
(3.26) 



cot A + cotB 
Check 

E^cotB - 1) + E fl (cot4 + 1) + N^cot B + 1) + 

+ N^cotA - 1) - (E c + N c )(cot4 + cot B) = 

(3.27) 

Using the values of Example 3.13, 

Bearing AB = tan-' - 727-64/+ 2815-36 = N 14° 29' 28" W 

= 345° 30' 32" 

Bearing AC = 278° 13' 57" .'• Angle A = 67° 16' 35" 

Bearing BC = 182° 27' 44" 

Bearing BA = 165° 30' 32" ••• Angle B = 16° 57' 12" 

Bearing CB = 02° 27' 44" 

Bearing CA = 098° 13' 57" .'• Angle C = 95° 46' 13" 



cot A = 0-41879 
From equation (3.25), 



CO-ORDINATES 151 

Check 1 = 180° 00' 00" 

cotB = 3*280 40 cotC = -0-10105 

P *L A cot B + E B cot A - /SNab 
& c = 



cot .4 + cotB 

(13486-85 x 3-2804) + (12759-21 x 0-41879) - 2815-36 

0-418 79 + 3-28040 
44 242-26 + 5343-43 - 2815-36 



E c = 12643-40 
From equation (3.26), 



3-699 19 



N^, = 



N,4 cot B + N B cot A + A E AB 



cot A + cotB 
(10 327-36x3-2804) + (13142-72x0-41879) - 727-64 

3-699 19 
33877-87 + 5504-04 - 727-64 



3-699 19 



N c = 10 449-39 



Check (equation 3.27) 

E/cotB-l) = 13486-85 x 2-28040 = 30755-41 

E^cot^ + l) = 12759-21 x 1-41879 = 18102-64 

N^(cotB + l) = 10 327-36 x 4-280 40 = 44 205-23 

N B (cotA - 1) = 13142-72 x -0-58121 = - 7638-68 

= 85 424-60 
(E<? + N c Xcot^ + cotS) = (12643-40 + 10449 -39)(3-69919) 

= 85424-62 
Example 3.14 Given the co-ordinates of four stations, 
B C 



A tL zoO-00 N 100-00 

B E 320-70 N 170-70 

C E 520-70 N 170-70 

D E 652-45 S 263*12 




Fig. 3.24 



to find the co-ordirtates of the intersection of the lines AC and BD. 



152 SURVEYING PROBLEMS AND SOLUTIONS 

Method 1 



AE 

tan bearing BD = -r__ 

AN 



652-45 - 320-70 331-75 



-263-12 - 170-70 -433-82 
bearingSD = S37°24'E = 142° 36' = bearingfiX 

250-0 - 520-7 -270-7 



tan bearing CA = 



100-0 - 170-7 -70-7 



bearing CA = S75°22'W = 255° 22' = bearing CX 
In triangle BCX 

BC = 520-7 - 320-7 = 200 (No difference in latitude, therefore 

due E) 

Bearing BC = 090° 

BX = 142° 36' 

.". Angle XBC = 52° 36' 

Bearing CB = 270° 

CX = 255° 22' 

Angle BCX = 14° 38' 

Length BX = BC sinBCX = 200 sin 14° 38' cosec (52° 36'+ 14° 38') 
sin BXC 

logBX = 1-73875 

To find co-ordinates of X. (Length BX known. Bearing S37°24'E) 
Logs 



; 1-522 21 > 


+ 33-28 (AE) 


E4 


+ 320-7 


sin bearing 1*78346 




AE 


33-28 


length 1-73875 




E* 


353-98 


cos bearing 1-90005 








1-63880 > 


- 43-53 (AN) 


N* 


+ 170-70 






AN 


- 43-53 



Ans. X = E 353-98 N 127-17 N* + 127-17 



Method 2 

From the previous method, 



tan bearing BD (/S) = Zj&k = -°' 76472 

tan bearing CA (a) = ~ ' ' = 3-82885 

-70-7 



CO-ORDINATES 153 

Using equation (3.18), 

AN = AE CB - AN gg tanj8 

tan a - tan )S 
AE^ = AN cx tana 

E N 

B 320-70 170-70 

C 520-70 170-70 

AE CB - 200-00 AN CB 0-0 

tana - tan0 = 3-828 85 + 0-76472 = 4-59357 

AN ~ - ?Hf - ° - - 43 - 538 

N c = 170-70 

N x = 127-16 

AE CX = -43-538 tana 

= -43-538x3-82885 = -166-70 

E c = 520-70 

E^ = 354-00 

Method 3 

By normal co-ordinate geometry, 

the equation of line AC = ^ ~ y i = y z ~ ^i 

i.e. y-100 = 170-7 - 100 = 70j7 = 

x-250 520-7-250 270-7 uzolz 

y _100 = 0-261 2 (x - 250) (1) 

Similarly, 

the equation of line BD = * ~ 170 ' 7 = -263-12 - 170-7 

x - 320-7 652-45 - 320-7 

= ~ 433 ' 82 = -1-3076 
331-75 

i-e. y- 170-7 = - 1-3076 (x - 320-7) (2) 

Subtracting (1) from (2), 

70-7 = 1-5688 x - (250 x 0-2612) - (320-7 x 1-3076) 
= 1-568 8 x _ 65-3 - 419-343 2 

x = 555 ' 3432 = 354-00 
1-5688 



154 SURVEYING PROBLEMS AND SOLUTIONS 

Substituting in equation (1), 

y = 0-261 2 (x- 250) + 100 

= 0-2612(354-250) + 100 

= (0-2612 xl04) + 100 

= 127-16 

Ans. X = E 354-0 N 127-16 

N.B. All these methods are mathematically sound but the first has the 
advantages that (1) no formulae are required beyond the solution of 
triangles, (2) additional information is derived which might be re- 
quired in setting-out processes. 

Example 3.15. Equalisation of a boundary line. The following sur- 
vey notes refer to a boundary traverse and stations A and E are situ- 
ated on the boundary. 

Horizontal length (ft) 

253-2 

426-4 

543-8 

1260-2 

It is proposed to replace the boundary ABCDE by a boundary AXE 

where AX is a straight line and X is situated on the line DE. 

Calculate the distance EX which will give equalisation of areas 

on each side of the new boundary. 

(M.Q.B./S) 

Computation of co-ordinates with A as the origin, Fig. 3.25 
Line AB N 83° 14' E 253-2 ft 



Line 


Bearing 


AB 


N 83° 14' E 


BC 


S 46°30'E 


CD 


N 36°13'E 


DE 


S 23°54'E 





Logs 


E, 0-0 


AE 


2-400 42 
9-99696 


+ 251-44 


sin# 


E B + 251-44 


length 


2-40346 




cos 6 


9-071 24 


N A 0-0 


AN 


1-47460 


+ 29-83 



No + 29-83 



Line BC S46°30'E 426-4 



E B + 251-44 
AE 2-490 38 -» + 309-30 



sin0 9-86056 



CO-ORDINATES 



155 



length 2-62982 
cos 6 9-83781 
AN 2-46763 



N 5 + 29-83 
- 293-51 



N c - 263-68 



Line CD N 36° 13' E 543-8 



E c + 560-74 
AE 2-50691 -> + 321-30 
sin0 9-77147 E D + 882-04 
length 2-73544 
cos# 9-90676 N c - 263-68 



AN 



2-642 20 



_+ 438-74 
H D + 175-06 



Fig. 3.25 



-200 



-400 



-600 



■800- 



-1000 




200 400 800 800 1000 1200 1400 



By construction, 
Join BD. 

Draw line parallel to BD through C to cut ED at C, . 

Area of triangle BDC X = area of triangle BDC (triangles on same base and 
between same parallels). 
Join C X A. 

Draw line parallel to C,/4 through B to cut ED at 8,. 
Area of triangle AB^C = area of triangle ABC^. 

•'• Line AB X (X) equalises the irregular boundary in such a way that 
triangle ABP + triangle QDB^ = triangle PQC 

Length EX = 977.84 ft (calc.) = 978 ft (scaled). 



156 SURVEYING PROBLEMS AND SOLUTIONS 

Line DE S 23° 54' E 1260-2 











E D + 882-04 








AE 2-70805 


+ 510-57 








sin0 9-60761 
length 3-100 44 


E E + 1392-61 














costf 9-96107 


N^ + 175-06 








AN 3-06151 


- 1152-15 












N^ - 977-09 






Checks AE 


+ 251-44 


AN + 29-83 - 293-51 








+ 309-30 




+ 438-74 -1152-15 








+ 321-30 




+ 468-57 -1445-66 






2AE 


+ 510-57 
+ 1392-61 




+ 468-57 






SAN - 977-09 




Area of figure ABCDEA (see 
ft) (2) 


Chapter 11) 

(3) (4) 


(5) 




N 


E 


F.dep. B.dep. 


(3) (4) 


A 


0-0 


0-0 


+ 


251-44 +1392-61 


-1141-17 


B 


+ 29-83 


+ 251-44 


+ 


560-74 0-0 


+ 560-74 


C 


- 263-68 


+ 560-74 


+ 


882-04 + 251-44 


+ 630-60 


D 


+ 175-06 


+ 882-04 


+ 


L392-61 + 560-74 


+ 831-87 


E 


- 977-09 


+ 1392-61 




0-0 + 882-04 


- 882-04 



Double areas (1) x (5) 



A 

B 167 26-8 

C 166 276-0 

D 145627-0 

E 861832-0 

+ 1024185-8 
- 166 276-0 
2; 857 909-8 
428 954-4 sq ft 



From co-ordinates 



Bearing EA = tan~ 1 - 1 ^ 92 ' 61 = N 54° 56' 44" W 
= 305° 03' 16" 



CO-ORDINATES 



157 



- ' A A I' 



Length EA = 1392-61 cosee 54° 56' 44' 

= 1701-2 ft (x) 

Bearing ED = N 23° 54' 00" W = 336° 06' 00" 

Angle AED = 336° 06' 00" - 305° 03' 16" 

= 31° 02' 44" 

To find length EX (a) such that the area of triangle AXE is equal to 
428 954-4 sq.ft. 

Area of triangle AXE = ^ax sin AED 

- 2 area triangle AXE 
x sin AED 

2 x 428 954-4 

1701-2 sin31°02' 44" 

= 977-84 ft (length EX). 

Exercises 3(c) (Boundaries) 

14. The undernoted bearings and measurements define an irregular 
boundary line on a mine plan between two points A and B, the latter 
being a point on a straight line XY , bearing from South to North. 

Plot the bearings and measurements to a scale of 1/2 in. = 100 ft, 
and thereafter lay down a straight line from A to a point on XBY so 
that the areas to the North and South respectively of that line will be 
equal. 

From A N 63° 30' W 185 ft 

S45°00'W 245 ft 

S80°45'W 175 ft 

N 55° 15' W 250 ft 

S 60° 30' W 300 ft to B 

Check your answer by calculation of the respective areas. 

(M.Q.B./M) 

15. The undernoted traverse was taken along an irregular boundary 
between two properties: 

AB N32°45'E 464 ft 

BC N71°30'E 308 ft 

CD S61°15'E 528 ft 

DE N71°30'E 212 ft 

EF S40°30'E 248 ft 

A lies on a straight boundary fence XY which bears N 7° 30' W. 



158 



SURVEYING PROBLEMS AND SOLUTIONS 



Plot the traverse to a scale 1/2400 and thereafter set out a 
straight line boundary from F to a point G on the fence XAY so that 
the areas North and South of the line are equal. 

What length of fencing will be required ? 

How far is G from A ? 

(N.R.C.T. Ans. 1480 ft; AG 515 ft) 

3.6 Transposition of Grid 




New axis 



Fig. 3. 26 Transposition of grid 

Let the line AB (Fig. 3.26) based upon an existing co-ordinate 
system have a bearing 6 and length s. 

Then AE^ = s sin (9 

^AB = s COS0 

The co-ordinate system is now to be changed so that the origin of 
the new system is 0. 

The co-ordinates of the position of 'slew' or rotation 

A = E^ 

and the axes are rotated clockwise through an angle +a to give a new 
bearing of AB. 



i.e. /8 = - a 
or a = 6 - /8 , 



i.e. Old bearing - New bearing (3.28) 



The new co-ordinates of B may now be computed: 
E B = Ejj + s sin/S 

= E A + s sin 6 cos a - s cos 6 sin a 

Efl = ^A + &&AB cos a - &^AB sin a 



(3.29) 



Similarly, 



Nfl = NJ^ + s cos/8 



CO-ORDINATES 



159 



then, 



and 



= N A + s cos 6 cos a + s sin 6 sin a 
N'b = N^ + AN^ cos a + AE^sina (3.30) 

If a scale factor k is required (e.g. to convert feet into metres), 



E* 



= E 'a + AE^ 

= E^ + ZctAE^gCos a - AN^ sin a] 



K = Ni + ANJS 



i 5 = l^ + iiiMfl 

= N^ + /ctAN^ cos a + AE^sina] 



From the above, 

AE^ B = fctAE^ cos a - AN^ sin a] 

= mAE^ - nAN^ 
AN^ S = mAN^ + nAE AB 
where m = k cos a and n = k sin a 

If the angle of rotation (a) is very small, the equations 
plified as cosa-»0 and sin a -> a radians. 

E5 = V A + k[AE AB -, fiN AB a] 
N B = N^ + tfAN^ + AE^ a] 

Example 3.16 Transposition of grid 



(3.31) 
(3.32) 



(3.33) 
(3.34) 



are sim- 



(3.35) 
(3.36) 




Fig. 3. 27 



160 



SURVEYING PROBLEMS AND SOLUTIONS 



In Fig. 3.27, 


AE 


AN 


Let OA = 045° i.e. N45°E 400 


+ 282-84 


+ 282-84 


OB = 120° S60°E 350 


+ 303-10 


- 175-00 


OC = 210° S30°W 350 


-175-00 


-303-10 


OD = 330° N30°W 400 


-200-00 


+ 346-40 


If the axes are now rotated through -15 c 


' the bearings 


will be in 


creased by +15°. 


AE' 


AN' 


OA' = 060° N60°E 400 


+ 346-40 


+ 200-00 


OB' = 135° S45°E 350 


+ 247-49 


-247-49 


OC' = 225° S45°W 350 


+ 247-49 


-247-49 


OD' = 345° N15°W 400 


-103-52 


+ 386-36 


Applying the transposition of the grid formulae; 




equation (3.29) AE' = AEcosa- 


AN sina 




equation(3.30) AN' = AN cos a + 


AE sin a 




AE An Ae cosa AN sina An 


cosa Ae sina 


Ae' 



OA 
OB 
OC 
OD 



+ 282*84 
+ 303.10 
- 175-00 
-200-00 



+ 282-84 
-175-00 
-303.10 
+ 346-40 



+273-20 
+ 292-77 
-169-04 
-193-19 



-73-20 
+ 45-29 
+ 78-45 
-89-68 



+ 273-20 
- 169-04 
-292-77 
+ 334-60 



-73-20 
-78-45 
+45-29 
+ 51-76 



+ 346-40 
+ 247-48 
-247-49 
-103-51 



An' 

+ 200-00 
-247-49 
-247-48 
+ 386-36 



N.B. (1) cos (-15°) - +0-965 93 

(2) sin (-15°) = -0-258 82 

(3) If the point of rotation (slew) had a co-ordinate value (EqNq) 
based on the new axes, these values would be added to the par- 
tial values, AE'AN' to give the new co-ordinate values. 

3.7 The National Grid Reference System 

Based on the Davidson Committee's recommendations, all British 
Ordnance Survey Maps will, on complete revision, be based on the 
National Grid Reference System with the metre as the unit. 

The origin of the 'Modified Transverse Mercator Projection' for 
the British Isles is 

Latitude 49° N 
Longitude 2° W 

To provide positive co-ordinates for the reference system a 'False 
Origin' was produced by moving the origin 100 km North and 400 km 
West . 

The basic grid is founded upon a 100 km square; commencing from 
the false origin which lies to the S.W. of the British Isles, and all 
squares are referenced by relation to this corner of the square. 



CO-ORDINATES 



161 




/0 100 200 300 400 500 600 700km E 

False 

origin _1_ 49° N latitude 

True -^T 

2* W long'itucle 



True 
origin 



Fig. 3.28 Old O.S. grid reference system 



162 



SURVEYING PROBLEMS AND SOLUTIONS 



'Eastings are always quoted first.' 

Originally the 100 km squares were given a reference based on the 
number of 100 kilometres East and North from the origin (see Fig. 3.28). 

Subsequently, 500 km squares were given prefix letters of S, N and 
H, and then each square was given a letter of the alphabet (neglecting 
I). To the right of the large squares the next letter in the alphabet 
gives the appropriate prefixes, T, O and J (see Fig. 3.29). 



km ► 








1 


Central 
meridian 






L 


M 


N 


o 


P 


L 


M 


H- Hebrides 


Q 


R 


S 


T 


U 


Q 


1 R J 








V 


W 


X 


Y 


z 


V 


W 




A 


B 


C 


D 


E 


A 


B 




F 


G 


H 


J 


K 


F 


G 




■ 


800 
N- North 


L 


M 




O 


P 


L 


M 




Q 


R 


s S 


J T 


U 


Q 


L-*- 




V 


W 


X 


Y 


z 


V 


w 




A 


B 


c 


D 


E 


A 


B 




F 


G 


H 


J 


K 


F 


G 






-ir 


300 
S- South 


L 


M 


-N- 


\ O 


P 


L 


T 
II 






Q 


.' 






U 


a 


u 

R 








S 







V 


W 


X 


Y 


z 


V 


. 



/ 

False origin 



-?#— 49* N latitude 
True -^"^ < . 

origin -^ 2 W longitude 



Fig. 3.29 New O.S. grid reference system 

Square 32 becomes SO 
43 becomes SK 

17 becomes NM 

The basic reference map is to the scale 1/25000 (i.e. approxi- 
mately 2y 2 inches to 1 mile), Fig. 3.30. 

Each map is prefixed by the reference letters followed by two 
digits representing the reference numbers of the SW corner of the 
sheet. See example (Fig. 3.30), i.e. SK54. This shows the relation- 
ship between the various scaled maps and the manner in which each 
sheet is referenced. 



CO-ORDINATES 



163 



A point P in Nottingham Regional College of Technology has the 
grid co-ordinates E 457 076-32 m, N 340 224-19 m. Its full 'Grid Refer- 
ence' to the nearest metre is written as SK/5740/076 224 and the 
sheets on which it will appear are: 

Reference Scale 

SK 54 1/25 000 ( ^ 2*/ 2 in. to 1 mile) 

SK 54 SE 1/10 560 ( 6 in. to 1 mile) 

SK 57 40 1/2 500 ( ^ 25 in. to 1 mile) 

SK 57 40 SW 1/1 250 ( 2; 50 in. to 1 mile) 



Sheet size 


Grid size 


10 km 


1km 


5 km 


1km 


1km 


100 m 


500 m 


100 m 



10km 



350 000 



49 



48 



46 
45 



340000 



SK 5740 SW 
(1/1250) 



5km 



SK 54 
(1/25000) 



SK 54 SE 
(6" to imile) 



SK 5740 
(1/2500) 



$5 -* "» u>A oi O) noo» <o o» 

o o o 

8 ° 

o Fig. 3.30 O.S. sheet sizes 8 

Exercises 3(d) (Co-ordinates) 

16. The co-ordinates of stations A and B are as follows: 

Latitude Departure 

A +8257 m + 1321m 

S +7542 m - 146 m 

Calculate the length and bearing of AB (Ans. 244° 01'; 1632 m) 

17. The co-ordinates of two points A and B are given as: 

A N 188 -6 m E 922-4 m 
B S 495-4 m E 58-6 m 



164 SURVEYING PROBLEMS AND SOLUTIONS 

Calculate the co-ordinates of a point P midway between A and B. 

(Ans. S 153-4 m, E 490-5 m) 

18. The bearings of a traverse have been referred to the magnetic 
meridian at the initial station A and the total co-ordinates of B, rela- 
tive to A, are found to be 368 m W, 796 m S. 

Calculate (a) the length and magnetic bearing of AB, 

(b) the true bearing of AB assuming that the magnetic 
declination is 13° 10' W of true north, 

(c) the co-ordinates of B with reference to the true 
meridian at the initial station. 

(Ans. (a) AB mag. bearing 204° 49' 877 m 

(b) true bearing 191° 39'; 

(c) corrected co-ordinates 177-1 W, 858-9 S) 

19. The co-ordinates, in metres of two points, X and Y, are as 
follows: X W 582-47 m N 1279-80 m 

Y E 1191-85 m S 755-18 m 

Calculate the length and bearing of XY. 

(Ans. 2699-92 m; 138°54'50") 

20. Survey station X has a Northing 424-4 ft, Easting 213*7 ft and 
a height above Ordnance datum of 260*8 ft. 

Station Y has a Northing 1728-6 ft, Easting 9263-4 ft and a depth 
below Ordnance datum 763*2 ft. 

Find the length, bearing and inclination of a line joining XY . 

(Ans. 9143-3 ft; 081° 47' 54"; 1 is 8-92) 

21. The co-ordinates of four survey stations are given below: 

Station North (ft) East (ft) 

A 718 90 

B 822 469 

C 164 614 

D 210 81 
Calculate the co-ordinates of the intersection of the lines AC and 

BD. (L.U. Ans. N 520, E 277) 

22. Readings of lengths and whole circle bearings from a traverse 
carried out by a chain and theodolite reading to 1 minute of arc were as 
follows, after adjusting the angles: 

Line AB BC CD DE 

W.C.B. 0°00' 35° 40' 46° 15y 2 ' 156° 13' 

Length (ft) 487-2 538*6 448-9 295-4 

Line EF FA AG GH HD 

W.C.B. 180° 00' 270° 00' 64° 58' 346° 25' 37° 40' 

Length (ft) 963-9 756-2 459-3 590-7 589-0 



CO-ORDINATES 165 

Taking the direction AB as north, calculate the latitude and de- 
parture of each line. If A is taken as origin and the mean co-ordinates 
of D as obtained by the three routes are taken as correct, find the co- 
ordinates of the other points by correcting along each line in propor- 
tion to chainage (answers are required correct to the nearest 0*1 ft) 

(L.U. Ans. D = 1234-7 ft N, 637-6 ft E) 

23. The following notes were taken during a theodolite traverse: 
Bearing of line AB 14° 48' 00" 

Angle observed Length (metres) 

ABC 198° 06' 30" AB 245 

BCD 284° 01 '30" BC 310 

CDE 200° 12' 30" CD 480 

DEF 271° 33' 30" DE 709 

EFG 268° 01 '30" EF 430 

FG 607 
Calculate the length and bearing of the line GA. 

(Ans. 220-6 m; N 61° 27' 40" W) 

24. From the following notes, calculate the length and bearing of the 
line DA: 

Line Bearing Length 

AB 015° 30' 630 m 

BC 103° 45' 540 m 

CD 270° 00' 227 m 

(Ans. 668 m; S 44° 13' W) 

25. The notes of an underground traverse in a level seam are as 
follows: 

Line Azimuth Distance (ft) 

AB 30° 42' 

BC 86° 24' 150-6 

CD 32° 30' 168-3 

DE 315° 06' 45-0 

The roadway DE is to be continued on its present bearing to a 
point F such that F is on the same line as AB produced. 
Calculate the lengths of EF and FB. 

(M.Q.B./M Ans. EF 88-9 ft; FB 286-2 ft) 

26. A shaft is sunk to a certain seam in which the workings to the 
dip have reached a level DE. It is proposed to deepen the shaft and 
connect the point E in the dip workings to a point X by a cross- 
measures drift, dipping at 1 in 200 towards X. The point X is to be 



Line 


Azimuth 


AB 


270° 00' 


BC 


184° 30' 


CD 


159° 15' 


DE 


90° 00' 



166 SURVEYING PROBLEMS AND SOLUTIONS 

134 ft from the centre of the shaft A and due East from it, AX being 
level. 

The following ate the notes of a traverse made in the seam from 
the centre of the shaft A to the point E. 

Distance Vertical Angle 

127 Level 

550 Dipping 21° 

730 Dipping 18y 2 ° 

83 Level 

Calculate (a) the azimuth and horizontal length of the drift EX 
and (b) the amount by which it is necessary to deepen the 
shaft. 

(M.Q.B./M Ans. (a) 358° 40' 1159-6 ft 
(b) 434-5 ft) 

27. The notes of a traverse between two points A and £ in a certain 
seam are as follows: 

Distance (ft) Angle of Inclination 

600 +6° 

450 -30° 

550 level 

355 level 

It is proposed to drive a cross-measures drift from a point E to 
another point F exactly midway between A and B. 
Calculate the azimuth and length EF. 

(M.Q.B./M Ans. 359° 33'; 867 ft, 888-3 ft inclined) 

28. Undernoted are details of a short traverse between the faces of 
two advancing headings, BA and DE, which are to be driven forward 
until they meet: 

Line Azimuth Distance 

AB 80° 270-6 ft 

BC 180° 488-0 ft 

CD 240° 377-0 ft 

DE 350° 318-0 ft 

Calculate the distance still to be driven in each heading. 

(M.Q.B./M Ans. BA + 168-4 ft; DE + 291-5 ft) 

29. In order to set out the curve connecting two straights of a road 
to be constructed, the co-ordinates on the National Grid of /, the point 
of intersection of the centre lines of the straights produced, are re- 
quired. 



L,ine 


Azimuth 


AB 


89° 


BC 


170° 


CD 


181° 


DE 


280° 



CO-ORDINATES 



167 



A is a point on the centre line of one straight, the bearing Al 
being 72° 00' 00", and B is a point on the centre line of the other 
straight, the bearing IB being 49° 26' 00" . 

Using the following data, calculate with full checks the co-ordinates 
of/. 

Eastings (ft) Northings (ft) 

A +43758-32 +52202-50 

B +45165-97 +52874-50 

The length AB is 1559-83 ft and the bearing 64° 28' 50". 

(N.U. Ans. E + 45 309-72 N + 52706-58) 

30. It is proposed to sink a vertical shaft to connect X on a roadway 
CD in the upper horizon with a roadway GH in the lower horizon which 
passes under CD. From surveys in the two horizons the following data 
are compiled: 



Upper horizon 

Station Horizontal Angle 


Inclination 


Inclined 
Length 


Remarks 


A 






co-ordinates of A 




+ 1 in 200 


854-37 


E 6549-10 ft 


B 276° 15' 45" 






N 1356-24 ft 




+ 1 in 400 


943-21 


Bearing AB 


C 88° 19' 10" 






N 30° 14' 00" E. 



Level 



D 



Lower horizon 

Station Horizontal Angle Inclination 



736-21 



Inclined 
Length 



+ lin50 326-17 



G 



H 



193° 46' 45" 
83° 03' 10" 



+ 1 in 20 



level 



278-66 



626-10 



Remarks 

Co-ordinates of £ 
E 7704-08 ft 
N 1210-88 ft 
Bearing EF 
N 54° 59' 10" E. 



Calculate the co-ordinates of X (Ans. E 8005*54 ft, N 1918*79 ft) 

31 . The surface levels of two shafts X and Y and their depth are 

respectively as follows: 

Surface Level Depth 

X 820-5 ft A.O.D. 200yd 

Y 535*5 ft A.O.D. 150 yd 



168 SURVEYING PROBLEMS AND SOLUTIONS 

The co-ordinates of the centre of the two shafts in fact, are re- 
spectively as follows: 

E N 

X -778-45 +2195-43 

Y +821*55 + 359-13 

Calculate the length and gradient of a cross-measures drift to con- 
nect the bottom of the shaft. 

(Ans. 2439-3 ft(incl.), 2435-6 ft(hor.); 3° 10', i.e. 1 in 18) 

32. The co-ordinates of A are N25m E13m. From ,4 a line AB runs 
S44°ll' E for 117m. On the line AB an equilateral triangle ABC is 
set out with C to the north of AB. 

Calculate the co-ordinates of B and C. 

(Ans. B E + 94-5m, N-55-9m. C E+ 126-4 m, N + 53-7m) 

33. (a) Calculate the gradient (as a percentage) between two points, 
M and N, which have been co-ordinated and heighted as given below: 

p • t Co-ordinates Height 

E(ft) N(ft) (ft) 

M 6206-5 3465-2 212-4 

# 5103-2 2146-8 196-6 

6002-5 2961-4 

(b) Determine the length (in centimetres) of the line MN when 
plotted at a scale 1 : 500 (assume 1 f t = 0-3048 m). 

(c) Calculate the bearing of the line MO 

(R.I.C.S. Ans. 0-92%; 104-8cm;202°03') 

34. From an underground traverse between 2 shaft wires A and D 
the following partial co-ordinates in feet were obtained: 

AB E 150-632 ft, S 327 -958 ft 

BC E 528-314 ft, N 82-115 ft 

CD E 26-075 ft, N 428-862 ft 

Transform the above partials to give the total Grid co-ordinates of 
station B given that the Grid co-ordinates of A and D were: 

A E 520 163-462 metres, N 432 182-684 metres 

D E 520378-827 metres, N 432238-359 metres 

(Aide memoire) X = x, + K(x -yd) 

Y = y, + K(y + xd) 

(N.R.C.T.) 

Bibliography 

MINISTRY OF DEFENCE, Textbook of Topographical Surveying, 4th ed. 
(H.M.S.O.) 



COORDINATES 



169 



GLENDENNING, J., Principles of Surveying (Blackie) 
CLARK, D., Plane and Geodetic Surveying, Vol. 1 (Constable) 
HOLLAND, J.L., WARDELL, K. and WEBSTER, A. C, Surveying, Coal- 
mining Series, (Virtue) 

A Brief Description of the National Grid and Reference System. O.S. 
Booklet No. 1/45 (H.M.S.O.) 

middleton and CHADWICK, Treatise on Surveying, Vol. 1 (Spon). 
salmon, V.G., Practical Surveying and Field Work (Griffin) 
BRINKER, R.C. and TAYLOR, W.C., Elementary Surveying, 4th ed. 
(International Textbook Co.) 



Appendix 



Comparison of scales 



Scales in common use with the 








metric system 














Scales in common use with the 
foot/inch system 


Recommended 


Other Alternative 


byBSI 


Scales 








i : ioooooo 


\ 


1 


ioooooo 


A in to 1 mile approx. 


I : 500000 


1 : 625000 


1 


: 625000 


Id in to 1 mile approx. 


1 : 200000 


1 : 250000 


1 


: 250000 


i in to 1 mile approx. 


1 : 100000 


1 : 125000 


1 


: 126720 


\ in to 1 mile 


1 : 50000 


1 : 62500 


1 


: 63360 


1 in to 1 mile 


1 : 20000 


1 : 25000 


1 


: 25000 


2 J in to 1 mile approx. 


1 : 10000 




1 


: 10560 


6 in to 1 mile 


1 : 5000 










1 : 2000 


1 : 2500 


1 


: 2500 


1 in to 208.33 ft 


1 : 1000 


1 : 1250 


1 


: 1250 


1 in to 104.17 ft 


1 : 500 




1 


: 500 


1 in to 41.6 ft 






1 


:384 


A in to r ft 


1 : 200 




1 


: 192 


■is in to 1 ft 


1 : 100 




1 


: 9 6 


J in to 1 ft 


1 :so 




1 


: 4 8 


i in to 1 ft 


1 : 20 




1 


:24 


\ in to 1 ft 


1 : 10 




1 


: 12 


1 in to 1 ft 


1 :5 




1 


= 4 


3 in to 1 ft 



From Chartered Surveyor, March, 1968. 



4 INSTRUMENTAL OPTICS 



4.1 Reflection at Plane Surfaces 
4.11 Laws of reflection 

(1) The incident ray, the reflected ray, and the normal to the 
mirror at the point of incidence all lie in the same plane. 

(2) The angle of incidence (i) = the angle of reflection (r). 

A ... 8 




C 

Fig. 4.1 

The ray AO, Fig. 4. 1, is inclined at a (glancing angle MO A) to 
the minor MN. Since i = r, angle BON = MO A = a. If AO is pro- 
duced to C, 

Angle MO A = NOC = BON = a 

Thus the deviation of the ray AO is 2a. Therefore the deviation 
angle is twice the glancing angle, 

i.e. D = 2a (4.1) 

4.12 Deviation by successive reflections on two inclined mirrors 
(Fig. 4.2) 

Ray AB is incident on mirror A^/V, at a glancing angle a. It is 
thus deflected by reflection + 2a . 

The reflected ray BC incident upon mirror M 2 N Z at a glancing 
angle j8 is deflected by reflection - 2j8 (here clockwise is assumed 
+ve). 

170 



INSTRUMENTAL OPTICS 171 



\ 



Fig. 4.2 
The total deflection D is thus (2a- 2/3) = 2(a-j8). 

In triangle BCX, )8 = a + 

=0-a 

i.e. D = 2(a-j8) = 20 (4.2) 

As is constant, f/ie deflection after two successive reflections 
is constant and equal to twice the angle between the mirrors. 

4.13 The optical square (Fig. 4.3) 

This instrument, used for setting out right angles, employs the 
above principle. 

By Eq. (4.2), the deviation of any ray from 2 incident on mirror M 2 
at an angle a to the normal = 20, i.e. 2 x 45° = 90°. 

4.14 Deviation by rotating the mirror (Fig. 4.4) 

Let the incident ray AO be constant, with a glancing angle a. 
The mirror M, N, is then rotated by an anticlockwise angle /3 to MgN^. 

When the glancing angle is a the deviation angle is 2a. 

After rotation the glancing angle is (a+ /3) and the deviation angle 
is therefore 2(a+ /S ) . 

Thus the reflected ray is rotated by 

<f> = 2(a+£)-2a = 2/8 (4.3) 



172 



SURVEYING PROBLEMS AND SOLUTIONS 



// the incident ray remains constant the reflected ray deviates by twice 
the angular rotation of the mirror. 




Fig. 4.3 Optical square 




Fig. 4.4 
4.15 The sextant 
Principles of the sextant (Figs. 4.5 and 4.6) 

Mirror M, , silvered, is connected to a pointer P. As M, is rotated 
the pointer moves along the graduated arc. 

Mirror M 2 is only half-silvered and is fixed. 

When the reading at P is zero, Fig. 4.5, the image K, reflected 
from both mirrors, should be seen simultaneously with K through the 
plain glass part of M 2 . With a suitable object K as the horizon, the 
mirrors should be parallel. 



INSTRUMENTAL OPTICS 



173 




From triangle M y EM z , 
from triangle M^QM 2 , 



Fig. 4.5 Zero setting on the sextant 

When observing an elevated object S, Fig. 4.6, above the horizon 
K, the mirror M, is rotated through angle <j> until S is simultaneously 
observed with K. The angle being measured is therefore 6. 

2a = 2/3 + 6 
6 = 2(a-0) 
90 + a = 90 + p + <f> 
a = j8 + <j6 

4 = a-j8 = £0 (4.4) 

i.e. f/ie rotation of the mirror is half the angle of elevation. 

When the mirrors are parallel, <f> = with the index pointer MP at 
zero on the graduated arm. When the angle 6 is being observed, the 
mirror is turned through angle 0, but the recorded value MP 2 = 6, 
i.e. 2<f>. 

N.B. (1) Horizontal angles are only measured if the objects are at 
the same height relative to the observer, which means that in most 
cases the angle measured is in an inclined plane. The horizontal 
angle may be computed from the equation (see page 107): 

cos - cos a, cos a. 



cos H.A. = 



i 

sin a, sina 2 



where 6 = the measured angle in the inclined plane and a, and a 2 
= vertical angles. 

(2) Vertical angles must be measured relative to a true or an 
artificial horizon. 



174 



SURVEYING PROBLEMS AND SOLUTIONS 

s 




Recorded value 8 



Fig. 4.6 Principles of the sextant 

4.16 Use of the true horizon 

(a) As the angle of deviation, after two successive reflections, is 
independent of the angle of incidence on the first mirror, the object 
will continue to be seen on the horizon no matter how much the 
observer moves. Once the mirror W, has been set, the angle between 
the mirrors is set, and the observed angle recorded. 

This is the main advantage of the sextant as a hand instrument, 
particularly in marine and aerial navigation where the observer's posi- 
tion is unstable. 

(b) If the observer is well above the horizon, a correction 50 is 
required for the dip of the horizon, Fig. 4.8. 



INSTRUMENTAL OPTICS 



175 




\ 

Fig. 4. 7 Box sextant 

The Nautical Almanac contains tables for the correction factor 86 
due to the dip of the horizon based on the equation: 

86 = -0-97 yjh minutes 

where h = height in feet above sea level, 

or 86 = -1-756 V^ minutes 

where H = height in metres above sea level. 



(4.5) 




Fig. 4.8 Dip of the horizon 



4.17 Artificial horizon (Fig. 4.9) 

On land, no true horizon is possible, so an 'artificial horizon' is 
employed. This consists essentially of a trough of mercury, the surface 
of which assumes a horizontal plane forming a mirror. 



176 



SURVEYING PROBLEMS AND SOLUTIONS 




£-|> 



^^J Mercury plane 



Fig. 4.9 Artificial horizon 
The vertical angle observed between the object S and the reflec- 
tion of the image S, in the mercury is twice the angle of altitude (a) 
required. 

Observed angle = S,ES = 2a 
True altitude = MS, S = a 

Rays SE and SS^ are assumed parallel due to the distance of S from 
the instrument. 



4.18 Images in plane mirrors 




Image 



Fig. 4. 10 Images in plane mirrors 

Object in front of the mirror is seen at E as though it were 
situated at /, Fig. 4.10. 

From the glancing angles a and /S it can be seen that 

(a) triangles OFC and ICF are congruent, 

(b) triangles OFD and IDF are congruent. 

Thus the point / (image) is the same perpendicular distance from the 
mirror as (object), i.e. OF = FI. 



INSTRUMENTAL OPTICS 



177 



4.19 Virtual and real images 

As above, the rays reflected from the mirror appear to pass through 
/, the image thus being unreal or virtual. For the image to be real, the 
object would have to be virtual. 

The real test is whether the image can be received on a screen: if 
it can be— it is real, if not— it is virtual. 

4.2 Refraction at Plane Surfaces 

Normal 



/= Angle of incidence 
r» Angle of refraction 



Boundary of media 




Refracted ray 



Fig. 4.11 

The incident ray AO, meeting the boundary between two media, 
e.g. air and glass, is refracted to B, Fig. 4.11. 



4.21 Laws of refraction 

(1) The incident ray, the refracted ray, and the normal to the bound- 
ary plane between the two media at the point of incidence all lie in the 
same plane. 

(2) For any two given media the ratio — . is a constant known as 

sin r 

the refractive index (the light assumed to be monochromatic). 

sin i 



Thus 



Refractive Index = 



sin r 



(4.6) 



4.22 Total internal reflection (Fig. 4.12) 

If a ray AO is incident on a glass/ air boundary the ray may be 
refracted or reflected according to the angle of incidence. 



178 



SURVEYING PROBLEMS AND SOLUTIONS 




(b) 
Critical angle 



(C) 
Reflected 



Fig. 4.12 

When the angle of refraction is 90°, the critical angle of incidence 

is reached, i.e. 

sin c . /a n\ 

rMa = TTTono = sin c (4.7) 



g' 



= sin c 
1-5 



For crown glass the refractive index a fi g 
•' sinc-^g 

c ^ 41° 30' 

If the angle of incidence (glass/ air) i : > 41° 30' , the ray will be 
internally reflected, and this principle is employed in optical prisms 
within such surveying instruments as optical squares, reflecting prisms 
in binoculars, telescopes and optical scale-reading theodolites. 
N.B. Total internal reflection can only occur when light travels from 
one medium to an optically less dense medium, e.g. glass/ air. 

4.23 Relationships between refractive indices (Fig. 4.13) 

(a) // the refractive index from air to glass is a fj, g , then the re- 
fractive index from glass to air is g fM a 

Therefore e fjL a = 

e.g., if a /J. g = 1-5 (taking air as 1), then 



(4.8) 



8 



u„ = — = 0-66 
^ a 1-5 _ 



(b) Given parallel boundaries of air, glass, air, then 
sin i = constant 
sin i a 



(4.9) 



/** = 



-Ma = 



sin i 



8 



sin i 



sin i, 



aPi 



gfia sin i a = a fi g sin i g = constant 



INSTRUMENTAL OPTICS 



179 




Fig. 4.13 

(c) The emergent ray is parallel to the incident ray when returning 
to the same medium although there is relative displacement. 
This factor is used in the parallel plate micrometer. 



4.24 Refraction through triangular prisms 

When the two refractive surfaces are not parallel the ray may be 
bent twice in the same direction, thus deviating from its former direc- 
tion by an angle D. 

It can be seen from Fig. 4.14 that 

A = ft + ft (4.10) 

and D = (a, - ft) + (cl,- ft) (4.11) 

= (a 1+ a 2 )-(ft + ft) (4.12) 

i.e. D = (a, + a 2 ) - A (4.13) 

Thus the minimum deviation occurs when a, + Og = A (4.14) 

If A is small, then 

a = /x/3 

and D = /ift + /xft -A 

= Mft+ft)-^ 

= AQi-1) (4.15) 



180 



SURVEYING PROBLEMS AND SOLUTIONS 




Fig. 4.14 Refraction through a triangular prism 



4.25 Instruments using refraction through prisms 

The line ranger (Fig. 4.15) 
0^ 

Prism P 2 




Prism P^ 




Field of view 



Fig. 4.15 The line ranger 

a+)8 = 90° 
.'. 2(a+ /3) = 180° 

Thus 0, C0 2 is a straight line. 



INSTRUMENTAL OPTICS 



181 



The prism square (Fig. 4.16) 



^-~* 




Fig. 4.16 The prism square 

This is precisely the same mathematically as the optical square 
(Fig. 4.3), but light is internally reflected, the incident ray being 
greater than the critical angle of the glass. 

The double prismatic square (Fig. 4.17) combines the advantages of 
both the above hand instruments. 



Fig. 4.17 




182 



SURVEYING PROBLEMS AND SOLUTIONS 



Images 2 and 3 are reflected through the prisms. 0, is seen 
above and below the prisms. 

The parallel plate micrometer (Fig. 4.18) 

t 




Fig. 4.18 The parallel plate micrometer 

A parallel- sided disc of glass of refractive index /a and thickness 
t is rotated through an angle 6. Light is refracted to produce displace- 
ment of the line of sight by an amount x. 

x = DB = AB sin(0 - <f>) 

= — — : x sin(0 - d>) 
cos 

f(sinfl cos<ft - cosfl sin<ft) 
~ cos<£ 

= f(sin# - cos0 tan</>) 



but refractive index 

fj. = 



sin0 = 



sin# 

sin 4> 

sinfl 



and cos<£ = y/\l - sin 2 <£} 
vV 2 -sin 2 0} 



-<[ 



INSTRUMENTAL OPTICS 
cos0 sin# 1 



183 



sin0 - 



= t sin0 
= t sin 6 



1- 



VOtx 2 - sin 2 0). 
cos0 



VO? -sin 2 0) 



If is small, then sin0~0 rad and sin 2 may be neglected. 

.-. x * ifl(l-i) (4.17) 

Example 4.1 A parallel plate micrometer attached to a level is to 
show a displacement of 0*0 1 when rotated through 15° on either side 
of the vertical. 

Calculate the thickness of glass required if its refractive index is 
1-6. 

State also the staff reading to the nearest thousandth of a foot when 
the micrometer is brought to division 7 in sighting the next lower read- 
ing of 4*24, the divisions running to 20 with 10 for the normal posi- 
tion. (L.U.) 

Using the formula 

x = t sin 



'Hfc£h)\ 



t = 



siae h . // (l-s^)(l + si,^ 1 
[ *J \(jjl - sin 0)Qi + sin &))} 
0-01 x 12 



in. 



sin 



icoN _ If (l-sinl5)(l + sinl5) )] 
L a/ 1(1-6 -sin 15) (1-6+ sin 15)/ J 

0-01 x 12 
0-25882 x 0-388 24 in ' 
= 1-1940 in. 
N.B. If the approximation formula is used, t = 1-222 in. 

The micrometer is geared to the parallel plate and must be corre- 
lated. Precise levelling staves are usually graduated in feet and 
fiftieths of a foot, so the micrometer is also divided into 20 parts, 
each representing 0*001 ft. (The metric staff requires a metric micro- 
meter). 

To avoid confusion, the micrometer should be set to zero before each 
sight is taken and the micrometer reading is then added to the staff reading 
as the parallel plate refracts the line of sight to the next lower reading. 



184 



SURVEYING PROBLEMS AND SOLUTIONS 



Z* 






a 






3 





ZP 






!R 












- • 






-» 

- s 
-* 






\ 





Reading 4-24 Staff 

0-0070 Micrometer 
4-2470 

Fig. 4.19 Use of the parallel plate micrometer in precise levelling 
Exercises 4(a) 

1. Describe the parallel plate micrometer and show how it is used in 
precise work when attached to a level. 

If an attachment of this type is to give a difference of 0-01 of a 
foot for a rotation of 20°, calculate the required thickness of glass 
when the refractive index is 1*6. 

Describe how the instrument may be graduated to read to 0-001 of 
a foot for displacements of 0-01 of a foot above and below the mean. 

(L.U. Ans. 0-88 in.) 

2. Describe the method of operation of a parallel plate micrometer in 
precise levelling. If the index of refraction from air to glass is 1*6 and 
the parallel plate prism is 0*6 in. thick, calculate the angular rotation 
of the prism to give a vertical displacement of the image of 0*001 ft. 

(L.U. Ans. 3° 03' 36") 

4.3 Spherical Mirrors 

4.31 Concave or converging mirrors (Fig. 4.20) 

A narrow beam of light produces a real principal focus F. P is 
called the pole of the mirror and C is the centre of curvature. PF is 
the focal length of the mirror. 

A ray AB, parallel to the axis, will be reflected to F. BC will be 
normal to the curve at B, so that 



Angle ABC = 
As AB is parallel to PC, 

Angle PCB = 
BF = 



angle CBF = 6. 

angle ABC = 6 
FC. 



INSTRUMENTAL OPTICS 



185 




Fig. 4.20 Concave mirror 

As the beam is assumed narrow 

PF ^ BF ~ FC 

:. PC ^ 2PF = 2/ ^ r. (4.18) 

If the beam of light is wide a cusp surface is produced with the 
apex at the principal focus. The parabolic mirror overcomes this ano- 
maly and is used as a reflector for car headlights, fires, etc, with the 
light or heat source at the focus. 




Fig. 4.21 Wide beam on a circular mirror 



186 



SURVEYING PROBLEMS AND SOLUTIONS 




Fig. 4.22 Parabolic mirror 
4.32 Convex or diverging mirrors (Fig. 4.23) 




C^" 



Fig. 4.23 Convex mirror 

A narrow beam of light produces a virtual principal point F, 
being reflected away from the axis. 

The angular principles are the same as for a concave mirror and 

r e 2/ 

4.33 The relationship between object and image in curved 
mirrors 

Assuming a narrow beam, the following rays are considered in all 
cases (Fig. 4.24). 

(a) Ray OA, parallel to the principal axis, is reflected to pass 
through the focus F. 

(b) Ray OB, passing through the focus F, is then reflected 
parallel to the axis. 

(c) Ray OD, passing through the centre of curvature C, and thus 
a line normal to the curve. 

N.B. In graphical solutions, it is advantageous to exaggerate the 
vertical scale, the position of the image remaining in the true position. 
As the amount of curvature is distorted, it should be represented as a 



INSTRUMENTAL OPTICS 



187 



straight line perpendicular to the axis. 

Any two of the above rays produce, at their intersection, the posi- 
tion of the image /. 



\ 





























^ 


\^i* / 




'! 


^^r\. 


/C Oy 










1 








/*\ 




l 







! 

% Fig. 4.24 





The relationships between object and image for concave mirrors 
are: 

(a) When the object is at infinity, the image is small, real, and 
inverted. 

(b) When the object is at the centre of curvature C, the image is 
also at C, real, of the same size and inverted. 

(c) When the object is between C and F, the image is real, en- 
larged and inverted. 

(d) When the object is at F, the image is at infinity. 

(e) When the object is between F and P, the image is virtual, en- 
larged and erect. 

For convex mirrors, in all cases the image is virtual, diminished 
and erect, Fig. 4.25. 




Fig. 4.25 



4.34 Sign convention 



There are several sign conventions but here the convention Real- 
is-positive is adopted. This has many advantages provided the work 
is not too advanced. 



188 



SURVEYING PROBLEMS AND SOLUTIONS 



All real distances are treated as positive values whilst virtual 
distances are treated as negative values— in all cases distances are 
measured from the pole. 

N.B. In the diagrams real distances are shown as solid lines whilst 
virtual distances are dotted. 

4.35 Derivation of Formulae 

Concave mirror (image real), Fig. 4.26 



Object „ 



c r v b 




f r 1 


|, 


u 1 


1 2 

/ r " 


/ 

Fig. 4.26 

1 1 

= -+ - 
U V 



To prove: 

where / = the focal length of the mirror 
r = the radius of curvature 
u = the distance of the object from the pole P 
v = the distance of the image from the pole P 
The ray OA is reflected at A to AI making an equal angle a on 
either side of the normal AC. 
From Fig. 4.26, 

6 = a+ fi .'. a = 6-13 

and (f> = 2a + j8 

= 209-j3) + i 8 
= 2B-P 
i.e. <£+j8= 26. 
As the angles a, /3, 6 and <f> are ^ small, B is closely adjacent to 



to P. 



<f>rad ~ Sin< £ = Jp 



P rad ■£! Sin j8 = 


OP 


6 . - sin0 = 

rad 


h 
CP 


h h 2h 
1P + 0P ~ CP 





I is real so IP is +ve. 
is real so OP is +ve. 



i.e. 



1 + 1 = 2 

v u r 



INSTRUMENTAL OPTICS 



■ J < as f=T ) 



189 
(4.19) 



Concave mirror (image virtual), Fig. 4.27 




Fig. 4.27 



From Fig. 4.27, 

= £ - a ••• a = j6 - 
and <f> = 2a - j8 

= 208-0-0 

= -20 + /8 
/. 20 = j8-0 



As before, 



2fc _ _ft _^_ 
CP~ OP /P 



i.e. 



u v 



but the image is virtual, therefore v is negative. 



2 11 1 

r ~ f u v 



Convex mirror, Fig. 4.28 




Fig. 4.28 
<f> = a + ••• a = <f> - 6 

2a = + /S 
.-. 2(<£-0) = </> + j3 
i.e. <f> = 2(9 + /S 

<£-j8 = 20 



(4.19) 



190 



SURVEYING PROBLEMS AND SOLUTIONS 



As before, 



Thus 



trad ~ Siti< f> 


h 
~ -IP 


(/ is virtual 


.*. IP is -ve.) 


Prad ~ sin/3 


h 
~ OP 


(0 is real 


OP is +ve). 


. ^ sin0 

rad 


h 
~ -PC 


(C is virtual 


••• PC is -ve). 



h _ h _ 2h 
IP OP -PC 



1 1 

i.e. ^- - — 

-v u 



2_ 
-r 

JL = _L 

r f 

Therefore, using the sign convention, the formula is common to 
both types of mirror in all cases. 



1 1 

— + — 
v u 



(4.19) 



4.36 Magnification in spherical mirrors (Fig. 4.29) 

u 




Fig. 4.29 Magnification in spherical mirrors 
IB is the image of OA. 

In the right-angled triangles OPO^ and /P/, the angle a is common, 
being the angles of incidence and of reflection, and therefore the tri- 
angles are similar. 

... A . /7 i (image size) /, P (v) 

Thus, magnification = 



00^ (object size) 0, P (w) 
m = — neglecting signs 



(4.20) 



INSTRUMENTAL OPTICS 



191 



Example 4.2 An object 1 in. high is placed on the principal axis 
20 in. from a concave mirror which has a radius of curvature of 15 in. 
Find the position, size and nature of the image. 
As the mirror is concave, 



/ = 



the object is real 
Substituting in Eq. (4.19), 



15. 
+ 20 



i-I+I 




J U V 




1 = 1-1 




V / u 




2 1 


1 


"15 20" 


12 



•'. v = 12 in. 
Thus the image is real (but will be inverted) as v is positive. 

Magnification v 

m = — 
u 

Size of image = 0*6 in. 



-S-- 



4.4 Refraction Through Thin Lenses 
4.41 Definitions 

(a) Types of lens 

Convex (converging), Fig. 4.30 (a) 
Concave (diverging), Fig. 4.30(b) 



Double convex 



Piano- convex 
(a) 



Convex meniscus 



Double concave Plano-concave Concave meniscus 

(b) 

Fig. 4.30 Types of lens 



192 SURVEYING PROBLEMS AND SOLUTIONS 

(b) Focal points (Fig. 4.31) 




Convex lens 




(pole) "1 Princ i pal axi s 



Concave lens 

Fig. 4.31 Conjugate foci 

4.42 Formation of images (Fig. 4.32) 

If a thin lens is assumed to be split into a series of small prisms, 

any ray incident on the face will be refracted and will deviate by an 

angle 

D = A(/x -1) (Eq.4.15) 




Fig. 4.32 Formation of images 
N.B. The deviation angle D is also related to the height h and the 
focal length /, i.e. 

D = h/f (4.21) 



INSTRUMENTAL OPTICS 



193 



4.43 The relationship between object and image in a thin lens 

The position of the image can be drawn using three rays, Fig. 4.33. 




Fig. 4. 33 

N.B. Two principal foci, F, and F 2 , exist. 

(a) Ray OA parallel to the principal axis is refracted to pass 
through principal focus F 2 . 

(b) Ray OB passes through the principal focus F, and is then 
refracted parallel to the principal axis. 

(c) Ray OPI passes from object to image through the pole P 
without refraction. 

Convex lens 

(a) When the object is at infinity, the image is at the principal 
focus F 2 , real and inverted. 

(b) When the object is between infinity and F, , the image is real 
and inverted. 

(c) When the object is between F, and P, the image is virtual, 
magnified, and erect, i.e. a simple magnifying glass. 

Concave lens 

The image is always virtual, erect and diminished. 

4.44 Derivation of formulae 

The real-is-positive sign convention is again adopted, but for con- 
vex lenses the real distances and focal lengths are considered posi- 
tive, whilst for concave lenses the virtual distances and focal lengths 
are considered negative. 

As with mirrors, thin lens formulae depend on small angle approxi- 
mations. 



Convex lens 

(a) Image real, Fig. 4.34 



D = a + (3 



194 



SURVEYING PROBLEMS AND SOLUTIONS 




By Eq. (4.21), 



D = 1 



1 = 1+ - 

U V 



.... i =± + L 

U V 



(b) Image virtual, i.e. object between F and P, Fig. 4.35. 




Fig. 4. 35 

D = a - /3 

h. = A_A 

/ U V 

but v is virtual, i .e. negative. 



i.e. 



Concave lens (Fig. 4.36) 



4 -i + i 

/ u v 



D = ft - a 



i.e. 



INSTRUMENTAL OPTICS 

f V u 

but v and / are negative, being virtual distances 

/ u v 



195 




Fig. 4. 36 

4.45 Magnification in thin lenses (Fig. 4.37) 




Fig. 4. 37 

As with spherical mirrors, OPO^ and /P/, are similar right-angled 
triangles with angle a commpn. 

//i (image size) 
OOi (object size) 
UP (image distance v) 



magnification m = _L 



y P (object distance u) 
— as before 



196 



SURVEYING PROBLEMS AND SOLUTIONS 



N.B. This should not be confused with angular magnification or magni- 
fying power (M), which is defined as 

the angle subtended at the eye by the image 
the angle subtended at the eye by the object 

For the astronomical telescope, with the image at infinity, 



M = 



focal length of objective _ jo 
focal length of eyepiece f e 



(4.22) 



4.5 Telescopes 
4.51 Kepler's astronomical telescope (Fig. 4.38) 



Objective 



Eyepiece 

Diaphragm 




Fig. 4.38 Kepler's atronomical telescope 

The telescope is designed to increase the angle subtending dis- 
tant objects and thus apparently to bring them nearer. 

The objective lens, converging and of long focal length, produces 
an image FX, inverted but real, of the object at infinity. 

The eyepiece lens, converging but of short focal length, is placed 
close to F so as to produce from the real object FX a virtual image 
IY , magnified but similarly inverted. 

4.52 Galileo's telescope (Fig. 4.39) 

The eyepiece is concave and produces a virtual, magnified, but 
erect image IY of the original inverted image XF produced by the 
objective. As the latter image lies outside the telescope eyepiece, it 
is unsuitable for surveying purposes where cross hairs are required. 



INSTRUMENTAL OPTICS 



197 



Objective 




Fig. 4.39 Galileo's telescope 

4.53 Eyepieces 

Ideally, the eyepieces should reduce chromatic and spherical 
aberration. 

Lenses of the same material are achromatic if their distance apart 
is equal to the average of their focal lengths, i.e. 

d = i(/,+/ 2 ) (4-23) 

If their distance apart is equal to the differences between their 
focal lengths, spherical aberration is reduced, i.e. 

d = /, - U (4.24) 

For surveying purposes the diaphragm must be between the eye- 
piece and the objective. The most suitable is Ramsden* s eyepiece, 
Fig. 4.40. 



Diaphragm 




Fig. 4.40 Ramsden's eyepiece 

The focal length of each lens is the same, namely /. Neither of the 

conditions (4.23) or (4.24) is satisfied. 

Chromatic |-(/i+/ 2 ) = / compared with 2/3/ 

Spherical /, - / 2 = compared with 2/3 / 

Huyghen's eyepiece, Fig. 4.41, satisfies the conditions but the 
focal plane lies between the lenses. It is used in the Galileo telescope. 



198 



SURVEYING PROBLEMS AND SOLUTIONS 



Diaphragm 




Fig. 4.41 Huyghen's eyepiece 
Chromatic condition -j (3/ + f) = 2/ = d 
Spherical condition 3/ - / = 2/ = d 

Example 4.3 An astronomical telescope consists of two thin lenses 
24 in. apart. If the magnifying power is xl2, what are the focal lengths 
of the two lenses ? 




Fig. 4.42 
magnifying power = 



= 12 



12/. = h 



But f + f e = 26 in. 

/. 12/ e + f e = 26 in. 
26 



f e = =-r - 2 in. eyepiece lens 



fo = 12/ e 



24 in. objective lens 



4.54 The internal focussing telescope (Fig. 4.43) 

The eyepiece and objective are fixed and an internal concave lens 
is used for focussing. 

For the convex lens, by Eq. (4.19) 



1 

/, 



1 + 1 



INSTRUMENTAL OPTICS 



199 



Objective 



Internal focussing 
lens 



Diaphragm 




i.e. 



For the concave lens, 



Fig. 4.43 


Internal focussing telescope 


1 1 


- 1 or 1 

u, u, 


1 1 


«2 


= -(f,-<0 




1 

/a 


.1+1 

"2 V 2 




1 

u 


1 

= + 

v, - d 


1 

l-d 


1 

/a 


1 
v, -d / 


1 
-d 



(4.25) 



An internal focussing telescope has a length / from the objective 
to the diaphragm. The respective focal lengths of the objective and the 
internal focussing lens are /, and / 2 . 

To find the distance d of the focussing lens from the objective 
when the object focussed is u, from the objective, Fig. 4.43. 

For the objective, \ 11 






For the focussing lens, 



/a 



-JU 



i.e. 



1 _ 1 



u. 



i.e. 



v, - d / - d 
(v, - d)(/ - d) = / a (/ - d) - / 2 (v, - d) 
d 2 - dtf + v,) + { / Vl -/ 2 (/-v,)} = 
d 2 - dC/ + v,) + W/ + / a ) -f 2 l\ = 



(4.26) 



200 



SURVEYING PROBLEMS AND SOLUTIONS 



This is a quadratic equation in d and its value will vary according to 
the distance «, of the object from the instrument. 

Example 4.4 Describe, with the aid of a sketch, the function of an 
internal focussing lens in a surveyors' telescope and state the advant- 
ages and disadvantages of internal focussing as compared with external 
focussing. 

In a telescope, the object glass of focal length 7 in. is located 9 
in. away from the diaphragm. The focussing lens is midway between 
these when the staff 60 ft away is focussed. Determine the focal length 
of the focussing lens. (L.U.) 



Internal focussing 
lens 



Diaphragm 




Fig. 4.44 

For the convex objective lens, 
/, = 7 in. 
m, = 60 x 12 = 

Then, by Eq. (4.19), 

1 - JL - i 

= 1- JL 

7 720 
720 - 7 



720 in. 



720 



x 



For the focussing lens, 

v, - 4-5 = 7-068 
4-5 
1 



u 2 
1 

u 



713 
5040 



4-5 = 2-568 



+ 



u. 



u - 



2-568 4-5 

-4-5 + 2-568 
11-556 

-5-98 in. 



(i.e. the lens is concave) 



INSTRUMENTAL OPTICS 



201 



Example 4.5 In an internally focussing telescope, Fig. 4.43, the 
objective of focal length 5 in. is 7*5 in. from the diaphragm. If the in- 
ternal focussing lens is of focal length 10 in., find its distance from 
the diaphragm when focussed to infinity. 

For the objective, /, = 5 in. and thus the position of F, will be 
5 in. from C,. 

C 2 F, = 5 - d 

For the internal focussing lens, 

/a = -10 

u 2 = -(5-d) 

v, = 7-5 - d 



i.e. 



i.e. 



1 = ± + J- 

f 2 U 2 *2 




1 - l . X 




10 5 _ d ' 7-5 - d 




-(5 - <0(7-5 - d) = -10(7-5 - d) + 10(5 - d) 




(37-5 - 12-5d + d z ) = -75 + lOd + 50 - lOd = 


-25 


d 2 - 12-5d + 12-5 = 




d = 4-235 in. 




v 9 = 7-5 - 4-235 = 3-265 in. 





i.e. the internal focussing lens will be 3-265 in. away from the dia- 
phragm when focussed to infinity. 



4.55. The tacheometric telescope (external focussing) (Fig. 4.45) 

Staff 



\fertical axis 



Diaphragm 




Fig. 4.45 The tacheometric telescope (external focussing) 

Let a, b and c represent the three horizontal cross hairs of the 
diaphragm, ac being a distance i apart and b midway between a and 



202 SURVEYING PROBLEMS AND SOLUTIONS 

With the telescope in focus, these lines will coincide with the 
image of the staff observed at A, B and C respectively; the distance 
AC = s is known as the staff intercept. The line bOB represents the 
line of collimation of the telescope, with bO and OB conjugate focal 
lengths of the lens, v and u, respectively. The principal focal length 
of the lens is FO (/), whilst the vertical axis is a distance k from 
the principal focus F. 

Because the triangles acO and ACO are similar, 

AC = QB or ± = u_ 
ac ob i v 

Using the lens formula, Eq.(4.19), 

f u v 
and multiplying both sides by uf gives, 

V 

Substituting the value of u/v from Eq.(4.27), 

u = s- + / 
i 

Thus the distance from the vertical axis to the staff is given as 

D = sl+ (/+d) (4.28) 

This is the formula which is applied for normal stadia observations 
with the telescope horizontal and the staff vertical. 

The ratio f/i - M is given a convenient value of, say, 100 (occa- 
sionally 50), whilst the additive constant (/ + d) = K will vary depend- 
ing upon the instrument. 

The formula may thus be simplified as 

D = M.s + K (4.29) 

Example 4.6 The constants M and K for a certain instrument were 
100 and 1*5 respectively. Readings taken on to the vertical staff were 
3*15, 4*26 and 5*37 ft respectively, the telescope being horizontal. 
Calculate the horizontal distance from the instrument to the staff. 

The stadia intercept s = 5-37-3-15 = 2-22 ft 
Horizontal distance D = 100 x 2-22 + 1-5 
= 223-5 ft (68-1 m) 

If the instrument was set at 103-62 ft A.O.D. and the height to the 
trunnion axis at 4-83 ft, 



INSTRUMENTAL OPTICS 



203 



then the reduced level of the staff station = 103-62 + 4-83 - 4-26 

= 104-19 ft A.O.D. 

(31-757 m) 

N.B. 4-26 - 3-15 = 5-37 - 4-26 = 1-11 = Is 

If the readings taken on to a metre staff were 0*960, 1*298, 1*636 
respectively, then the horizontal distance = 100 x (1*636-0*960) 
= 67*6m + 0*5 m = 68*1 m 

If the instrument was set at 31*583 m A.O.D. and the height of the 
trunnion axis at 1*472 m, 

then the reduced level of the staff station = 31*583 + 1*472 - 1-298 

= 31*757 m 



4.56. The anallatic lens (Fig. 4.46) 

Vertical axis 
Anallatic 
Diaphragm ■•*>* 



Staff 




Fig. 4.46 The anallatic lens 

In the equation D = s(f/i) + (/ + d), the additive factor (/ + d) 
can be eliminated by introducing a convex lens between the objective 
and the diaphragm. 

The basic principles can be seen in Fig. 4.46. The rays from the 
staff Ad and Ce will for a given distance D always form a constant 
angle d intersecting at G. If this fixed point G is made to fall on the 
vertical axis of the instrument the additive term will be eliminated. 

Consider the object lens with the object AC and the image a, c,, 
i.e. neglecting the anallatic lens. 



By Eq.(4.19) 



and by Eq. (4.27) 



1= 1+1 

f U V 

u _ s 
v a,c. 



(4.30) 



(4.31) 



Consider the anallatic lens with the object as a^c x and the image 



204 SURVEYING PROBLEMS AND SOLUTIONS 

as ac. Thus the object distance = v, - x and 
the image distance = v - x 
Applying the previous equations to this lens, 

i * _ * 

U ~ v i - * v - x 



(4.32) 



and v ' ~ * = J^L (4.33) 

v - x a x c t 

N.B. The object distance is assumed positive but the image distance 
is negative. 

An expression for D can now be found by eliminating v, v, and 

a^c^ from these four equations. 

From Eq. (4.32) 

v, - x = *< v ~ *> 
/, + v - * 

From Eq.(4.33) 

ac(v - x) 

a,Ci = 

v, - x 

Combining these gives 

a r - Q c (/i + v-x) 

"f 1 — 7 

but from Eq. (4.30) 

v = -Hi 

Substituting in the above 

a i c t = 
but from Eq. (4.31) 

a,c, = 

giving 

sf 



u - f 






ac[f, 


+ Uf - 
+ u-f 


«) 




U 




sv 
u 


sf 
u - f 




ac(fi 


+ uf - 

u-f 


') 



i.e. 



s//, = aciu - O (/, + ■£- - *) 

Writing ac as i, the distance apart of the stadia lines 
sfU = HfM-f) + uf - *(«-/)] 
= *[«(/,+/-*) + /(*-/,)] 

u = s //i - /(* ~ /t) 

»(/+/,-*) / + /, - x 



INSTRUMENTAL OPTICS 205 

but D = u + d 

t(/ + /,-*) f + U-x 
= Ms - ^ x ~ f'> + d 

and if d = ^-^ (4-34) 

f + U-x 

D = Ms (4.35) 



where M = 



//i 



i (/ + /,-*) (4.36) 

a constant factor usually 100. 

The manufacturer can therefore choose the lenses where the focal 
length / t is such that /, < x < f . 

Today, this is mainly of academic interest only, as all instruments 
have internal focussing telescopes, and the tacheometic formula D = 
/(s/i) + (f + d) is not applicable; nor can the internal focussing be 
considered anallatic as it is movable. 

The variation of the focal length of the objective system is gener- 
ally considered to be negligible for most practical purposes (see Ex- 
ample 4.7), manufacturers aiming at a low value for K, and in many 
cases the telescopes are so designed that when focussed at infinity 
the focussing lens is midway between the objective and the diaphragm. 
This allows accuracies for horizontal sights of up to 1/1000 for most 
distances required in this type of work. 

Example 4.7 An anallatic telescope is fitted with an object lens of 
6 in. focal length. If the stadia lines are 0*06 in. apart and the vertical 
axis 4 in. from the object lens, calculate the focal length of the anal- 
latic lens and its position relative to the vertical axis if the multiplying 
constant is 100. 

From Eq. (4.34) the distance between objective and axis 



= 4 





f+U-x 


When / = 6 in. 


d _ 6 (*-/i) 
6 + /, - x 


Also, from Eq.(4.36), 


M ~ & 


!(/+/, -X) 


Therefore when M = 100, 


i = 0-06, /=6, 


M - 


6/, 



0-06(6 + U-x) 



100 



206 SURVEYING PROBLEMS AND SOLUTIONS 

Combining these equations, 

4(6+/,-*) = 6(*-/,) 

10* = 24 + 10/, 

and 100 x 0-06(6 + /,-*) = 6/, 

6* = 36+0 

* = 6 in. 

and /, = 3*6 in. 

Thus the focal length of the anallatic lens is 3*6 in. and its posi- 
tion is (6-4) = 2 in. from the vertical axis. 

Example 4.7a An anallatic tacheometer in use on a remote survey 
was damaged and it was decided to use a glass diaphragm not origin- 
ally designed for the instrument. The spacing of the outer lines of the 
new diaphragm was 0*05 in., focal lengths of the object glass and the 
anallatic lens 3 in., fixed distance between object glass and trunnion 
axis 3 in., and the anallatic lens could be moved by an adjusting 
screw between its limiting positions 3 in. and 4 in. from the object 
glass. In order to make the multiplier 100 it was decided to adjust the 
position of the anallatic lens, or if this proved inadequate to graduate 
a special staff for use with the instrument. Make calculations to de- 
termine which course was necessary, and if a special staff is required, 
determine the correct calibration and the additive constant (if any). 
What is the obvious disadvantage to the use of such a special staff? 

(L.U.) 
From Eq. (4.36), ,, 

M = ILL 



JL 

Mi 



/ + u - u 



= 3 



100 x 0-05 

= 6 - 1-8 = 4-2 in. 

i.e. the anallatic lens should be 4*2 in. from the objective. As this is 

not possible, the lens is set as near as possible to this value, i.e. 

4 in. 

Then M = ^-*- 3 - = 90 

0-05(3 + 3 - 4) — 

The additive factor K from Eq.(4.34) 

= /(*-/,) = 3(4-3) = V5in 
/+/,-* 3 + 3-4 



INSTRUMENTAL OPTICS 



207 



If the multiplying factor is to be 100, then the staff must be graduated 
in such a way that in reading 1 foot the actual length on the staff is 



12 x 15 in. i.e. 13! in. 

9 3 



4.57. The tacheometric telescope (internal focussing) (Fig. 4.47) 

Mechanical axis 



Diaphragm 




H4 3£ ? 



u 2 



Fig. 4.47 Tacheometric telescope (internal focussing) 

To find the spacing of the stadia lines to give a multiplying fac- 
tor M for a given sight distance: 

if i = stadia interval, 

and s = stadia intercept; 

for the convex lens (objective) 



s u, 



i.e. 



where m, is the magnifying power, 
for the concave lens (focussing) 



* _ v 2 



= Vfir 



i.e. 



x = £Tl = m ,s 



i = xYl = m 2 x 
i = m^TTizS 



(4.37) 



but 



distance D = Ms 

- '-$ 

M 



(4.38) 



Example 4.8 An internally focussing telescope has an objective 6 in. 
from the diaphragm. The respective focal lengths of the objective and 
the internal focussing lens are 5 in and 10 in. Find the distance apart 
the stadia lines should be to have a multiplying factor of 100 for an 
observed distance of 500 ft. 



208 SURVEYING PROBLEMS AND SOLUTIONS 

At 500 ft the object will be 500 ft - 6/2 in. from the objective, 
i.e. «, = 500 x 12 - 3 = 5997 in 

v i = mi X ? = 5-004 2 in. 
5997 - 5 

From Eq.(4.26), 

d 2 - d(/ + v,) + {v,(/ +/ 2 ) - / 2 /} = 
i.e. d z - d(6 + v,) + {16v, - 60} = 

d = | [(6 + v ) ± VK6 + v,) 2 - 64 v, + 240}] 
= |[(6 + v) + V(6-v 1 )(46-v 1 )]. 

i.e. d = i[ 11-004 2 ± VCO'995 8 x 40-995 8)] 

= 2-308 in. 
= / _ d = 6 - 2-308 = 3-692 in 



v 



u 2 = v,-d = 5-004 2-2-308 - 2-696 in 

From Eq.(4.38), 

. _ Dm } m 2 _ Dv^V2 
M Mu x u 2 

500xl2x 5-0042x3-692 
100 x 5997 x 2-696 

= 0-068 56 in. 



Example 4.9 What errors will be introduced if the previous instrument 

is used for distances varying from 50 to 500 ft ? 

At 50 ft rrt 

«, = 50 x 12 - 3 = 597 in. 



597x5 = 2985 
597 - 5 592 



= 5-0422 in. 



Then, from Eq.(4.26), 

d = |[(6 + v,) + 4(6-^X46-^)] 

= |[ll-042 2 - VC0-9578 x 40-9578)] 

= 2-389 in. 
v 2 = 6 - 2-389 = 3-611 in. 



"2 



= 5-042 - 2-389 = 2-653 in. 



INSTRUMENTAL OPTICS 209 

The stadia intercept (s) = 0-068 56 x " 1 " 2 

v,v 2 

0-06856 x 597 x 2-653 
5-042 x 3-611 

= 5-9641 in. 

- 0-497 Oft 

The value should be 0-500 

error = 0-0030 ft 

representing Q-30 ft in 100 ft 

At 100 ft error = 0-27 ft 

200 ft error = 0-20 ft 

300 ft error = 0-09 ft 

400 ft error = 0-01 ft 

500 ft error = 0-00 ft 

Example 4.10 An internal focussing telescope has an object glass 
of 8 in. focal length. The distance between the object glass and the 
diaphragm is 10 in. When the telescope is at infinity focus, the inter- 
nal focussing lens is exactly midway between the objective and the 
diaphragm. Determine the focal length of the focussing lens. 

At infinity focus the optical centre of the focussing lens lies on 
the line joining the optical centre of the objective and the cross-hairs, 
but deviates laterally 0*001 in. from it when the telescope is focussed 
at 25 ft. Calculate the angular error in seconds due to this cause. 

(L.U.) 

With the telescope focussed at infinity, v, = /, 
For the focussing lens, 



1 




1 1 








u 




«2 V 2 










= 


1 


1 

l-d 








v, - d 








1 


1 










U-d 


l-d 








= 


1 

8 -5 


1 


2 
15 






10-5 




u 


= 


7-5 in. 


Focal length of focussing 


lens 



With focus at 25 ft (assuming 25 ft from object lens.) 
u, = 25 x 12 = 300 

v, = J£jL = 30 °x 8 = 8-2192 in. 
«, - /, 300 - 8 



210 



SURVEYING PROBLEMS AND SOLUTIONS 



From Eq. (4.26), 



d 2 - dO + Vt) + IV1O + /2) -/a/I 







i.e. d 2 - 18-219 2d + (143-836 - 75) = 

Solving for d, d = 5-348 



/-if, 



\\i 






H-* oooi M 

T 



x 

' x J X\ 



Fig. 4.48 

With focus at 25 ft the image would appear at x, neglecting the 
internal focussing lens, i.e. OX = v,. With the focussing lens moving 
off line, the line of sight is now EXJ Z and all images produced by the 
objective appear as on this line. 

The line of sight through the objective is thus displaced XX, in 
the length v,. 



To calculate XX,. 




XX, 


XE 


/,/ 2 


UE 


i.e. x = 


0-001 x (/ - v,) 
l-d 




0-001 x (10 - 8*219) 




10 - 5-348 


= 


0-001781 = o-OOO 391 in. 
4-552 


To calculate the angular error (5), 


tan 8 = Ml 
OX 


_ X 


8 = 206 265x0-000 391 9 . 8 nccondc 

Q.010 



4.6 Instrumental Errors in the Theodolite 
4.61 Eccentricity of the horizontal circle 
In Fig. 4.49, let 0, = vertical axis 



INSTRUMENTAL OPTICS 



211 



2 = Graduated circle axis 
0^0 2 = e = eccentricity 
Z A, ~ 0,4 2 = r 




Fig. 4.49 

If the graduated circle (Centre 2 ) is not concentric with the ver- 
tical axis (centre 0,) containing the readers A and B, the recorded 
value 6 will be in error by the angle a. 

As the instrument is rotated, the readers will successively occupy 
positions 4,6,, A Z B 2 , A 3 B 3 . 



a, 



= 0i-<£i 



tan- M 
E 

e sin <f> 



A Z E 



tan 



a, ~ 



r - e cos 
e sin</> 



(4.39) 



(4.40) 



Since e is small compared with r and as a is small, a rad ~ tan a 
Similarly, tan a 2 = e sin<f> 



a. 



r + e cos0 
e sin0 



(4.41) 
(4.40) 



If the readers are 180° apart, 4,0,8, is a straight line and the 
mean of the recorded values 6 give the true value of the angle <f>. 



212 SURVEYING PROBLEMS AND SOLUTIONS 

i.e. <f> = 0, - a, = Z + a 2 

•"' 2cf> = 0, + 2 as a, ~ a 2 

4> = §(0, + 2 ) (4.42) 

N.B. (1) On the line X 2 , a = 0. 

(2) At 90° to this line, a = maximum. 

(3) If the instrument has only one reader, the angle should be 
repeated by transitting the telescope and rotating anticlockwise, thus 
giving recorded values 180° from original values This is of particular 
importance with glass arc theodolites in which the graduated circle is 
of small radius. 

To determine the amount of eccentricity and index error on the 
horizontal circle: 

(1) Set index A to 0° and read displacement of index B from 
180°, i.e. 5,. 

(2) Set index B to 0° and read displacement of index A from 
180°, i.e. 8 2 . 

(3) Repeat these operations at a constant interval around the 
plate, i.e. zeros at multiples of 10°. 

If the readers A and B are diametrically opposed, let 5, = dis- 
placement of reader B, from 180°, Fig. 4.50. 
Index A y at 0°. 
Index B, at 180° - 8,, i.e. 180 - (2a + A). 



Reading fli 
180-6 

Index error \ 




Fig. 4.50 

Let § 2 = displacement of reader A z from 180°, Fig. 4.51, 

Index B 2 at 0. 

Index A 2 at 180 - 8 2 , i.e. 180 - (2a + A). 



INSTRUMENTAL OPTICS 



213 




Index error X 



Fig. 4.51 

If there is no eccentricity and A and B are 180° apart, then 5, = 
S 2 = 0. 

If there is eccentricity and A and B are 180° apart, then 5, = 8 Z = 
a constant 

If there is no eccentricity and A and B are not 180° apart, then 
+ 5, = -S 2 , i.e. equal, but opposite in sign. 

If there is eccentricity and A and B are not 180° apart, then 5, and 8 2 
will vary in magnitude as the zero setting is consecutively changed 
around the circle of centre 2f but their difference will remain con- 
stant. 

A plotting of the values using a different zero for each pair of in- 
dex settings will give the results shown in Fig. 4.52. 




360 



+x and -x are 180" apart 



Fig. 4.52 

4.62. The line of collimation not perpendicular to the trunnion axis 

Let the line of sight make an angle of 90° + € with the trunnion 
axis inclined at an angle a, Fig. 4.53. 



214 



SURVEYING PROBLEMS AND SOLUTIONS 




Vfcrtical axis 

Fig. 4. 53 Line of collimation not perpendicular to the trunnion axis 

The angular error Q in the horizontal plane due to the error e may 
be found by reference to Fig. 4.53. 



f = TY tan e 
YZ = TY cos a 
= tan € sec a 





tan0 = *I 
YZ 






tane = *X 
TY 


i.e. X] 


But 


TY = YZ sec a 


i.e. 


.-. 


tan0 = TYt * ne = 
TY cos a 


If and e are small, 




then 


6 - 


= € sec a 



(4.43) 



(4.44) 

If observations are made on the same face to two stations of eleva- 
tions a , and a 2 , then the error in the horizontal angle will be 

±(#i - 2 ) = ±tan -1 (tan £ sec a,) - tan"' (tan e sec a 2 ) (4.45) 

±(0,-0 2 ) 2: ±e(seca, -seca 2 ) (4.46) 

On changing face, the error will be of equal value but opposite in 
sign. Thus the mean of face left and face right eliminates the error due 
to collimation in azimuth. The sign of the angle, i.e. elevation of de- 
pression, is ignored in the equation. 

The extension of a straight line, Fig. 4.54. If this instrument is 
used to extend a straight line by transitting the telescope, the follow- 
ing conditions prevail * 

With the axis on the line TQ the line of sight will be 0A V To 
observe A, the instrument must be rotated through the angle e to give 
pointing (1) — the axis will be rotated through the same angle E to 

T,Qr 

On transitting the telescope the line of sight will be (180° - 2a) 
A,OB, = A0B 2 . B 2 is thus fixed -pointing (2). 

On changing face the process is repeated — pointing (3) — and then 



INSTRUMENTAL OPTICS 



215 




Fig. 4.54 

pointing (4) will give position S 4 . 

The angle B 2 0B A = 4e, but the mean position B will be the cor- 
rect extension of the line AO. 

The method of adjustment follows the above process, B 2 B 4 being 
measured on a horizontal scale. 

The collimation error may be corrected by moving the telescope 
graticule to read on B 3 , i.e. aB^. 

4.63 The trunnion axis not perpendicular to the vertical axis 
(Fig. 4.55) 

The trunnion (horizontal or transit) axis should be at right angles 
to the vertical axis; if the plate bubbles are centralised, the trunnion 
axis will not be horizontal if a trunnion axis error occurs. Thus the 
line of sight, on transitting, will sweep out a plane inclined to the ver- 
tical by an angle equal to the tilt of the trunnion axis. 




Fig. 4.55 Trunnion axis not perpendicular to the vertical axis 

If the instrument is in correct adjustment, the line of sight sweeps 
out the vertical plane ABCD, Fig. 4.55. 

If the trunnion axis is tilted by an angle e, the line of sight sweeps 



216 



SURVEYING PROBLEMS AND SOLUTIONS 



out the inclined plane ABEF. 

In Fig. 4.55, the line of sight is assumed to be AE. To correct for 
the tilt of the plane it is necessary to rotate the horizontal bearing of 
the line of sight by an angle 6, to bring it back to its correct position. 

EC = BC tan e 
ED ED 



Thus 



sin# 



AD 



tane 



tana tane 



(4.47) 



(4.48) 



ED 
i.e. sin# = tana tane 

and if 6 and e are small, then 

6 = e tan a 

where 6 = correction to the horizontal bearing 

e = trunnion axis error 
a = angle of inclination of sight 

On transitting the telescope, the inclination of the trunnion axis 
will be in the opposite direction but of equal magnitude. Thus the 
mean of face left and face right eliminates the error. 

Method of adjustment (Fig. 4.56) 

(1) Observe a highly elevated target A, e.g. a church spire. 

(2) With horizontal plates clamped, depress the telescope to 
observe a horizontal scale 8. 

(3) Change face and re-observe A. 

(4) As before, depress the telescope to observe the scale at C. 

(5) Rotate horizontally to D midway between B and C. 

(6) Elevate the telescope to the altitude of A. 

(7) Adjust the trunnion axis until A is observed. 

(8) On depressing the telescope, D should now be observed. 




Fig. 4.56 Field test of trunnion axis error 



INSTRUMENTAL OPTICS 



217 



4.64 Vertical axis not truly vertical (Fig. 4.57) 

If the instrument is in correct adjustment but the vertical axis is 
not truly vertical by an angle E, then the horizontal axis will not be 
truly horizontal by the same angle E. 

Thus the error in bearing due to this will be 



E tana 



(4.49) 



This is a variable error dependent on the direction of pointing 
relative to the direction of tilt of the vertical axis, and its effect is 
not eliminated on change of face, as the vertical axis does not change 
in position or inclination. 



Face left 




Horizontal line 



e 1/ 
t£^ i f / Face r ight 




Trunnion axis not perpendicular to the vertical axis 



Face left 




Horizontal lin e S \ V—. Fac e right 



Vertical axis 
unchanged on 
change of face 

Vertical axis not truly vertical 
Fig. 4.57 




In Fig. 4.58, the true horizontal angle (0) A x OB y = angle (</>) 
A£)B 2 -( c i)Ei tana, + (c 2 ) E z tan a 2 . 

Thus the error in pointing (0) is dependent on (1) the tilt of the 
axis E, which itself is dependent on the direction of pointing, vary- 
ing from maximum (£ ) when on the line of tilt of the vertical axis to 
zero when at 90° to this line, and (2) the angle of inclination of the 
line of sight. 

To measure the value of e and E a striding level is used, Fig. 
4.59. 



218 



SURVEYING PROBLEMS AND SOLUTIONS 




Fig. 4.58 




Bubble axis 



Trunnion axis 




Vertical axis 



Face right 



Vertical axis 



Fig. 4.59 

Let the vertical axis be inclined at an angle E 
Let the trunnion axis be inclined at an angle e 
Let the bubble be out of adjustment by an angle j3. 

Then face left 

tilt of the trunnion axis = E - e 
tilt of the bubble axis = E - j8 



face right 



tilt of the trunnion axis = E + e 
tilt of the bubble a^is = E + jS 



INSTRUMENTAL OPTICS 219 

Mean tilt of the trunnion axis = -|[(E - e) + (E + e)] = E (4.50) 

Mean tilt of the bubble axis = -±[(E - j8) + (E + j8)] = E (4.51) 

Therefore the mean correction taking all factors into account is 

Etana (4.52) 

N.B. The value of E is related to the direction of observation and its 
effective value will vary from maximum to nil. Tilting level readings 
should be taken for each pointing. 

If E is the maximum tilt of the axis in a given direction, then 

E, = E cos 6 

where is the angle between pointing and direction of maximum tilt. 

Then the bubble recording the tilt does not strictly need to be in 
adjustment, nor is it necessary to change it end for end as some 
authors suggest, the mean of face left and face right giving the true 
value. 

If the striding level is graduated from the centre outwards for n 
pointings and 2n readings of the bubble, then the correction to the 
mean observed direction is given by 

c = A (XL -1R) tana (4.53) 

2n 

where c = the correction in seconds 

d = the value of one division of the bubble in seconds 

XL = the sum of the readings of the left-hand end of the bubble 

2/? = the sum of the readings of the right-hand end of the bubble 

a = the angle of inclination of sight 

n = the number of pointings. 

The sign of the correction is positive as stated. Any changes de- 
pends upon the sign of XL - 2/2 and that of a. 

N.B. The greater the change in the value of a the greater the effect 
on the horizontal angle. 

4.65 Vertical circle index error (Fig. 4.60) 

When the telescope is horizontal, the altitude bubble should be 
central and the circle index reading zero (90° or 270° on whole circle 
reading instruments). 

If the true angle of altitude = a 

the recorded angles of altitude = a, and a 2 

the vertical collimation error = <f> 

and the circle index error = 6, 



220 



SURVEYING PROBLEMS AND SOLUTIONS 



Recorded value (F.L.) = a = a, - <f> - 6 
(F.R.) = a = a 2 + cf> + 
a = -1(0., + a 2 ) 



(4.54) 



Collimation error 



Bubble axis 



Eyepiece 
end 




Objective end 



Index error 






S 



Face left 





^\ / Collimation 
^X^" error 




& Ssf \ 
VC -''A «2 \ 




^fe.~r* -. -4 Index error 


ble axis— { ^J 


Mf 
^f^- 


\^s^ Face right 



Fig. 4.60 

Thus, provided the altitude bubble is centralised for each reading, 
the mean of face left and face right will give the true angle of altitude. 

If the bubble is not centralised then bubble error will occur, and, 
depending on the recorded displacement of the bubble at the objective 
and eyepiece ends, the sensitivity will indicate the angular error. As 
the bubble is rotated, the index is also rotated. 

Thus 6 will be subjected to an error of ± i"(0 - E)8", where 8" = 
the angular sensitivity of the bubble. 

If the objective end of the bubble is higher than the eyepiece end 
on face left, i.e. L > E L , then 6 will be decreased by ±-(0 L - B L )8", 
i.e. F.L. a = a, - <f> - {6 - \{0 L -E L )8"\ 



and 



F.R. a = a 2 + cf> + {0 + Uo R - E R )8"\ 



INSTRUMENTAL OPTICS 221 

a = |(a 1 + a 2 ) +y{(0 L+ /e ) ~(E L + E R )\ 

= i<a f + a 2 ) + |(20 - SB) (4.55) 

To *es* and adjust the index error 

(1) Centralise the altitude bubble and set the telescope to read 
zero (face left). 

(2) Observe a card on a vertical wall — record the line of sight 
at A. 

(3) Transit the telescope and repeat the operation. Record the 
line of sight at B. 

(4) Using the slow motion screw (vertical circle) observe the mid- 
point of AB. (The line of sight will now be horizontal.) 

(5) Bring the reading index to zero and then adjust the bubble to 
its midpoint. 

Example 4.11 Trunnion axis error. The following are the readings 
of the bubble ends A and B of a striding level which was placed on 
the trunnion axis of a theodolite and then reversed ('Left' indicates 
the left-hand side of the trunnion axis when looking along the tele- 
scope from the eyepiece end with the theodolite face right.) 

A on left 11-0, B on right 8*4 

B on left 10*8, A on right 8*6 

One division of the striding level corresponds to 15". All adjust- 
ments other than the horizontal trunnion axis adjustment of the theodo- 
lite being presumed correct, determine the true horizontal angle be- 
tween P and Q in the following observations (taken with the theodo- 
lite face left). 

Object Horizontal circle Vertical circle 

P 158° 20' 30" 42° 24' 

Q 218° 35' 42" 15° 42' 

(L.U.) 

By Eq.(4.53), 

Correction to bearing = -£-(%L - 2R) tana 

2n 

to P c = ^{(11-0 + 10-8) - (8-4 + 8-6)}tan 42° 24' 

= 15 x 4 ' 8 tan 42° 24' 
4 

= -18" tan 42° 24" = -16" 



222 



SURVEYING PROBLEMS AND SOLUTIONS 



to Q c = -18 tan 15° 42' = - 5' 

N.B. The correction would normally be positive when using the gen- 
eral notation, but the face is changed by the definition given in the 
problem. 

True bearing to P = 158° 20' 30" - 16" = 158° 20' 14" 

True bearing to Q = 218° 35' 42"- 5" = 218° 35' 37" 

True horizontal angle = 60° 15' 23" 

Example 4.12 In an underground traverse the following mean values 
were recorded from station B on to stations A and C 



A 
C 

Striding level: 



Station Horizontal Vertical 

Observed Angle Angle 

136° 21' 32" 

-13° 25' 20" 
+47° 36' 45" 

1 division = 10 seconds 
bubble graduated to 20 

Height of instrument at B 4-63 ft 

Height of target at A 3-42 ft 

at C 5-15 ft 

Ground length AB 256-32 ft 

BC 452-84 ft 

Calculate the gradient of the line AC 



Striding Level readings 



17-4 
14-5 



5-8 
2-7 



(R.I.C.S.) 



Striding level corrections 

to A L 17-4 

R 5-8 

2 )23-2 

11-6 

i.e. centre of bubble is 1-6 to left of centre of graduations. 

to B L 14-5 

R 2-7 

2 ) 17-2 

8-6 

i.e. centre of bubble is 1-4 to right of centre of graduations. 

The same, results may be obtained by using the basic equation 
(4.53) and transposing the readings as though the graduation were from 
the centre of the bubble. 



INSTRUMENTAL OPTICS 223 

i.e. to A 17-4 (L) becomes 7-4 (L) 

5-8 (R) becomes 4*2 (R) 

XL - XR 7-4 - 4-2 = 1#6 
2 2 

to C 14-5 (L) becomes 4-5 (L) 

2-7 (R) becomes 7-3 (R) 

XL - XR _ 4-5 - 7-3 = _ 1>4 
2 ~ 2 

Applying these values to Eq.(4.52), 

Correction to A = +1*6 x 10 x tan (-13° 25') 

= -3'8* 
Correction to C = -1-4 x 10 x tan(+47°37') 

= - 15,3 " 
Total angle correction = -11*5" 

Corrected horizontal angle = 136° 21' 32" - 11-5" 

= 136° 21' 20" 

To find true inclination of the ground and true distances (Fig. 4.61) 

Line AB Ba = sin" 1 1-21 cos 13° 25' 20" 

1 256-32 

5a, = -0°15'47" 
a, = 13° 25' 20" 
6 = 13° 09' 33" 
Horizontal length (D,) AB = 256-32 cos 13° 09' 33" = 249-59 ft 
Vertical difference (tf,) AB = 256-32 sin 13° 09' 33" = 58-35 ft 

Line BC . 0-52 cos 47° 36' 45" 
8a z = sin" 1 r=^rzrz 

2 452-84 

= -0°02'40" 

a 2 = 47° 36' 45" 

= 47° 34' 05" 

Horizontal length BC = 452-84 cos 47° 34' 05" = 305-54 ft 

Vertical difference = 452-84 sin 47° 34' 05" = 334-23 ft 

Difference in height AC = 58-35 + 334-23 = 392-58 ft 

To find the horizontal length AC : 
In triangle ABC, 

tan A -° 305-54 - 249-59 tan (180 - 136° 21' 20") 
2 305-54 + 249-59 2 



224 



SURVEYING PROBLEMS AND SOLUTIONS 



4-63 



4-63 



3-42 A 



TffKS^ 




5-15 



4-63 




4-63 



A-C 
A + C 



Fig. 4.61 



2° 18' 40" 



21°49'20' 



A = 24°08'00" 

Then AC = 305-54sinl36°21 , 20"cosec24°08 , 00" = 515-77 

Gradient AC = 392*58 ft in 515'77ft 

= 1 in 1-314 



INSTRUMENTAL OPTICS 225 

Example 4.13 (a) Show that when a pointing is made to an object 
which has a vertical angle h with a theodolite having its trunnion 
axis inclined at a small angle i to the horizontal, the error introduced 
into the horizontal circle reading as a result of the trunnion axis tilt 
is i tan/i. 

(b) The observations set out below have been taken at a station 
P with a theodolite, both circles of which have two index marks. On 
face left, the vertical circle nominally records 90° minus the angle of 
elevation. The plate bubble is mounted parallel to the trunnion axis 
and is graduated with the zero of the scale at the centre of the tube 
one division represents 20". 

The intersection of the telescope cross-hairs was set on signals 
A and B on both faces of the theodolite. The means of the readings 
of the circle and the plate bubble readings were : 

Signal Face Horizontal Circle Vertical Circle Midpoint of Bubble 
A 



B 



The vertical axis was then rotated so that the horizontal circle 
reading with the telescope in the face left position was 256° 40'; the 
reading of the midpoint of the bubble was then 0*4 division away from 
the circle. 

If the effect of collimation error c on a horizontal circle reading 
is c sec/i, calculate the collimation error, the tilt of the trunnion axis 
and the index error of the theodolite, the altitude of the vertical axis 
when the above observations were taken, and the value of the horizon- 
tal angle APB. (N.U.) 



At A (Fig. 4.62). As the bubble reading is equal and opposite, on 
change of face the horizontal plate is horizontal at 90° to the line of 
sight. The bubble is out of adjustment by 1 division = -20" F.L. 
At 90° to A the corrected bubble reading gives 

F.L. 0-4 + 1-0 = +1-4 div. = +28" 

The horizontal plate is thus inclined at 28" as is the vertical axis, in 
the direction A. 



left 


116° 39' 15" 


90° 00' 15" 


1-0 division 
towards circle 


right 


346° 39' 29" 


270° 00' 17" 


1-0 division 
towards circle 


left 


301° 18' 36" 


80° 03' 52" 


central 


right 


121° 18' 30" 


279° 56' 38" 


2-0 division 
towards circle 



226 



SURVEYING PROBLEMS AND SOLUTIONS 




Fig. 4.62 

At B (Fig. 4.63). The corrected bubble readings give 
F. 
F. 

N.B. This value may be checked (see p. 414) 



\L. 0-0+ 1-0 = +1-CH 
\R 2-0-1-0 = +1-0 J 



i.e. +20' 




166* (28") 



Fig. 4.63 



INSTRUMENTAL OPTICS 227 

Apparent dip = full dip cosine angle between 
= 28"cos(301-256) 
= 28" cos 45° 
= 20^_ 
The effect of instrumental errors in pointings (Fig. 4.64) 




Fig. 4.64 



Horizontal 

Collimation 6 C = ±c sec/i (if h = Q c = c); the mean of faces left 

and right gives the correct value. 

Trunnion Axis $ t = ± i tan ft (if h = d t = 0); the mean of faces left 
and right give the correct value. 

Vertical Axis d v = v tan/i (if h = 6 V = 0); the sign is dependent on 

the inclination of the axis. (F.L. inclination towards circle, with +h, 

d v is -ve 

Vertical 

Index error Bh = 2\h t -h r \; the mean of faces left and right gives the 

correct value 

Application to given values: 
At A (F.L.) 166° 39' 15" + c sech - v tank + i tan/i 
i.e. 166° 39' 15" + c 
(F.R.) 346° 39' 29" - c 
F.L. must equal F.R. 

.-. 166° 39' 15" + c = 346° 39' 29" - 180° - c 
2c = +14" 
c = + 7' (collimation error) 



228 SURVEYING PROBLEMS AND SOLUTIONS 

At B 

(F.L.) 301° 18'36" + 7 sec9°57" - 20 tan9°57' + i tan 9° 57" 

i.e. 301°18'36" +7-1 - 3-5 + 0-175* 
(F.R.) 121°18'30" -7-1 - 3-5 - 0-175 i 
F.L. must equal F.R. 

/. 301°18'36" + 7-1 -3-5 + 0-175 i = 121° 18' 30" + 180 - 7-1 - 3-5 - 

-0-175 i 
i.e. 0-35 i = -20-2" 

i = -58" (fli =-10-1") 

(trunnion axis error) 
Corrected readings 

F.L. A 166° 39' 15" + 7" = 166° 39' 22" 

B 301° 18' 36" + 7-1" - 3-5" - 10-1 = 301° 18' 29-5" 

Angle APB = 134° 39' 07-5" 
F.R. A 346° 39' 29" - 7" = 346° 39' 22" 

B 121° 18' 30" - 7-1 - 3-5 + 10-1 = 121° 18' 29-5" 

Angle APB = 134° 39' 07-5" 

Vertical angles 
At A: 8h = I{(90-90°00'15") - (270°00'17" -270)} = -16" 

At B: Sh = |{(90-80°03'52") - (279° 56' 38" -270)} = -15" 
N.B. The discrepancy is assumed to be an observational error. 

4.7 The Auxiliary Telescope 

This is used where steep sights are involved and in two possible 
forms: 

(1) Side telescope (2) Top telescope 

4.71 Side telescope 

There are two methods of using this form of telescope: (a) in adjust- 
ment and (b) out of adjustment with the main telescope. 

Adjustment 

(a) Alignment (Fig. 4.65). Observe a point A with the main tele- 
scope. Turn in azimuth to observe with the side telescope without 
altering the vertical circle. Raise or lower the side telescope until the 
horizontal cross-hair coincides with the target A. The horizontal hairs 
are now in the same plane. 



INSTRUMENTAL OPTICS 



229 




Main telescope 
Side telescope 



Fig. 4.65 Alignment of telescopes 




Fig. 4.66 

(b) Parallel lines of sight (Fig. 4.66). If x is the eccentricity of 
the telescope at the instrument this should be constant between the 
lines of sight (see Fig. 4.66). 

At a distance d x from the instrument, a scale set horizontally may 
be read as a,b, giving an intercept s,, and then at a\ readings a 2 b 2 
give intercept s 2 . 

If the lines of sight are parallel, 
If not, the angle of convergence/divergence e is given as 



e = tan 



-i s 2 



Si 



(4.56) 



d 2 — d, 

If s 2 > s,, the angle is +ve, i.e. diverging 

If s 2 < s,, the angle is -ve, i.e. converging. 

The amount of eccentricity x can be obtained from the same read- 
ings. 



230 



SURVEYING PROBLEMS AND SOLUTIONS 
d\ Si - X 



d 2 - d, 



i.e. 



x = Si(**2 - <*i) - ^1(^2 - Si) 
(d 2 -d,) 
s,d 2 — s 2 di 



(4.57) 
d 2 - d, 

By making the intercept s 2 = x, the collimation of the auxiliary 
telescope can be adjusted to give parallelism of the lines of sight. 

Observations with the side telescope 

(a) Vertical Angles. If the alignment is adjusted, then the true vertical 
angle will be observed. 

If an angular error of 8a exists between the main and the side 
telescope, then the mean of face left and face right observations is re- 
quired, i.e. 

F.L. = a, + da = a 
F.R. = a 2 - Sa = a 

a = i(a, + a 2 ) (4.58) 

(b) Horizontal Angles (Fig. 4.67) 




If 



Fig. 4.67 Horizontal angles with the side telescope 

6 = the true horizontal angle; 
<f> = the recorded horizontal angle; 
S, and 8 2 = errors due to eccentricity, 



INSTRUMENTAL OPTICS 231 

then = <t> x - 5, + 8 2 say F.L. 

6 = <f> 2 + S, - 8 2 F.R. 

i.e. = ya(^, + <^) (4.59) 

Example 4.14 In testing the eccentricity of a side telescope, read- 
ings were taken on to levelling staves placed horizontally at X and Y 
100 and 200 ft respectively from the instrument. 

Readings at X 5-42 ft 5-01 ft 

at Y 3-29 ft 2-79 ft 

Calculate (a) the collimation error (e), (b) the eccentricity (x). 

(R.I.C.S./M) 



From the readings, 


Si 


= 5-42 - 5-01 = 0-41 




s 2 


= 3-29 - 2-79 = 0-50 


Then, by Eq. (4.56), 










e 


- fan"' S2 ~ 


Si 




d 2 - 


<*, 




€" 


206265 x 
200 


(0-50 - 0-41) 
- 100 






= 185-6" = 


03' 06" 


by Eq. (4.57) 


X 


_ S\d 2 — s 2 d 
d 2 - d, 


i 






0-41 x 200 


- 0-50 x 100 






200 


- 100 






82 - 50 


0-32 ft 



Based on the metric system the question becomes: 

In testing the eccentricity of a side telescope, readings were taken 
on to levelling staves placed horizontally at X and Y, 30*48 m and 
60*96 m respectively from the instrument. 

Readings at X 1*652 m 1*527 m 

at y 1*003 m 0*850 m 

Calculate (a) the collimation error (€), (b) the eccentricity (x) 

s, = 1*652 - 1*527 = 0*125 m 

s 2 = 1*003 - 0*850 = 0*153 m 

Then e = 206 265 x (0*153-0*125) 

60*96 - 30*48 
= 189*5" = 03' 10" 



232 
and 



SURVEYING PROBLEMS AND SOLUTIONS 

0-125 X 60-96 - 0-153 x 30-48 
X ~~ 30-48 

= 0-097 m (0-32 ft) 



The effect of eccentricity x and collimation error e 

In Fig. 4.68, assuming small 4 

angles, 

Angle AOB = 8" 

206 265 x 



Angle BSC = e" 

206 265 y 



(4.60) 



(4.61) 



Angle AOC 



206265(x+y) (4 62) 
d 




Fig. 4.68 



e = 8 + e (4.63) 

••• Angle BOC = Angle BSC 

= £ 

As the eccentricity x is con- 
stant, the angle (5) is dependent 
upon the length of sight d. 

As the collimation angle € is 
constant, it has the same effect as 
the collimation error in the main 
telescope. It affects the horizontal 
angle by e sec a, where a is the vertical angle. 

Assuming the targets are at different altitudes, the true horizontal 
angle #,Fig. 4.67, is given as 

6 = <£, - (5, + €seca,) + (5 2 + eseca 2 ) say F.L. (4.64) 

Also 6 = <j> 2 + (8 2 + eseca,) - (5 2 + e seca 2 ) F.R. (4.65) 

6 = %(<£, + 2 ) (4.66) 

Thus the mean of the face left and face right values eliminates 
errors from all the above sources. 

Example 4.15 Using the instrument of Example 4.14, the following 
data were recorded: 



INSTRUMENTAL OPTICS 
(€ = 03' 06" x = 0-32 ft) 



233 



Station set 


Station 


Horizontal 


Vertical 


Remarks 


at 


observed 


circle 


circle 




B 


A 


0°05'20" 


+ 30° 26' 


Horizontal lengths 




C 


124° 10' 40" 


-10° 14' 


AB = 100' BC = 300' 

side telescope on 
right. 


Calculate the true horizontal angle ABC 












(R.I.C.S./M) 



By Eq. (4.64), 
True horizontal angle (6) = <f> - (5,+ e sec a,) + (5 2 + e secci^ ) 
206265 x 0-32 = 66 » 



5, = 



6\ = 



100 

206265 x 0-32 
300 



1'06' 



= 22" = 0'22" 



eseca, = 186" sec30°26' = 215-7", say 216" = 3' 36" 
eseca 2 = 186" sec 10° 14' = 189*0" = 3' 09" 

<f> = 124° 10' 40" - 0° 05' 20" = 124°05'20" 
6 = 124°05'20" - (1'06"+ 3'36") + (0' 22"+ 3' 09") 
= 124° 04' 09" 

4.72 Top telescope 

In this position the instrument can be used to measure horizontal 
angles only if it is in correct adjustment, as it is not possible to change 
face. 

Adjustment 

(a) Alignment. The adjustment is similar to that of the side tele- 
scope but observations are required by both telescopes on to a plumb 
line to ensure that the cross-hairs are in the same plane. 

(b) Parallel lines of sight (Fig. 4.69). Here readings are taken on 
vertical staves with the vertical circle reading zero. 

The calculations are the same as for the side telescope: 



6' = tan-' S z ~ s ' 
d, - d, 



(4.67) 



and 



Sid 2 — s 2 d, 

d 2 ~ dy 



(4.68) 



234 



SURVEYING PROBLEMS AND SOLUTIONS 







£' . 


" 




Si 




— -~. 


V 


• 


/ 


<*1 














<*2 



















Fig. 4.69 
Measurement of Vertical Angles (Fig. 4.70) 



Angle ATB is assumed 
equal to angle AOB as 
both are small values 




Fig. 4.70 Measurement of vertical angles with the top telescope 
The angular error (§') due to eccentricity is given as 



8' = tan-' 



(4.69) 



where 



d sec0 
x = eccentricity 
d = horizontal distance 
<£ = recorded vertical angle 

If e' is the error due to collimation then the true vertical angle 
(0) is given as 

±6 = ±(0+S'+ €') (4.70) 

assuming 8' and e' small. 

4.8 Angular Error due to Defective Centring of the Theodolite 

The angular error depends on the following, Fig. 4.72: 

(a) linear displacement x, 

(b) direction of the instrument B, relative to the station B 

(c) length of lines a and c. 



INSTRUMENTAL OPTICS 



235 



Locus of centring 
error 




Fig. 4.71 Minimum error due to defective centring of the theodolite 

The instrument may be set on the circumference of the circle of 
radius x. 

No error will occur if the instrument is set up at B, or B 2 (Fig. 
4.71), where A,B y ,B,B 2 and C lie on the arc of a circle. 




Fig. 4.72 The effects of centring errors 

In Fig. 4.72, let the instrument be set at B t instead of B. 
Angle 0, is measured instead of 6 
i.e. 6 = 0, - (a + B) 

Assume the misplumbing x to be in a direction <f> relative to the line 
AB. 



In triangle ABB U 



sin a = 



x sin<£ 
AB, 



(4.71) 



As the angle a is small, 

a _ 206265 x sin <f> 
c 



(4.72) 

Similarly, B" = 206265 x sin (0 - <^) (4 73) 

a 

.'. Total error E = a + B = 206265 Je^ + sin(fl - <ft) 1 (4.74) 

For maximum and minimum values, 

4*L = 206265 r[ cos< £ _ cos(fl - 0) 1 = Q 



236 SURVEYING PROBLEMS AND SOLUTIONS 

COS 4> _ COS (0 - (f>) 



i.e. 



c 



cos <ji = -£(cos 6 cos <£ + sin 6 sin </>) 



-j-sin<£ cot0 = £(cos0 cot<£ + sin0) 

cot0(l- ccos ^ = 



a 

c sin 6 



cot<£ = csind n (4.75) 

a - c cos d 

N.B. (1) If = 90°, 6 = or 180° 

(2) If a :» c, then <f> -* 90°, i.e. the maximum error exists 
when (f> tends towards 90° relative to the shorter line. 

(3) If a = c, <f> = 6/2. 

Professor Briggs proves that the probable error in the measured 
angle is 

, . ±2» //J: + I _!«££) (4.76) 

Example 4. 16 The centring error in setting up the theodolite at 
station B in an underground traverse survey is ±0*2 in. Compute the 
maximum and minimum errors in the measurement of the clockwise 
angle ABC induced by the centring error if the magnitude of the angle 
is approximately 120° and the length of the lines AB and BC is ap- 
proximately 80*1 and 79*8 ft respectively. 

(R.I.C.S.) 

(1) The minimum error as before will be nil. 

(2) The maximum error on the bisection of the angle ABC is AB ~ BC. 

i.e. <f> = = 60° a ~ c ~ 80 ft 

x = 9^1 = 0-0167 ft 
12 

E = 206265 x 0-016 7[ sin60 + sin(120-60) l 
L 80 80 J 

= 206265 x 0-0167 x 2 sin 60/80 

= 74 seconds i.e. l' 14" 

By Professor Briggs' equation, the probable error 

e = ± 2 x0-016 7 //_1_ J_ _ 2 cos 120 \ 
3-1416 aJ\S0 2 + 80 2 80+80 / 

= ±35" i.e. a 1/2 max error. 



INSTRUMENTAL OPTICS 237 

4.9 The Vernier 

This device for determining the decimal parts of a graduated scale 
may be of two types: 

(1) Direct reading 

(2) Retrograde 

both of which may be single or double. 



9 l 



(a) 



(b) 



| \ ■■■■■ *' | ' ■' ■ ' ■ ' I | . . . i | i i i i ) i i 

10 20 



® 5 

■i I i — i i i | — i v I j '| ' ' ' ' ' ' ' ' i » — i — i i I — r 



^ 



30 \ | 40 50 

Vernier scale 36-0 ■*• 0-3 =» 36-3 
reading 



I 5 © 

' ■'■ ' ■ | . '■ X \ N ■■ 



■ ■ ■ ■ i ■ ■ ' ■ r ■ ■ ■ ■ i ' • i 

{C) 30 3^.3 40 50 

Retrograde vernier 

Fig. 4.73 Verniers 

4.91 Direct reading vernier 

Let d = the smallest value on the main scale 
v = the smallest value on the vernier scale 
n = number of spaces on the vernier 
n vernier spaces occupy (ji - 1) main scale spaces 

i.e. nv = (n - l)d 

v = (n-l)d 
n 

Therefore the least count of the reading system is given by: 
d-v - d - «" - » 



= d 



d - v = d( n - n + *) = I 
\ n / n 



(4.77) 



(4.78) 



Thus the vernier enables the main scale to be read to -th of 1 division. 



238 SURVEYING PROBLEMS AND SOLUTIONS 

ii 
Example 4. 17 If the main scale value d = -^ and the number of 
spaces on the vernier (n) = 10, the vernier will read to 1/10 x 1/10 = 
1/100 in. 

4.92 Retrograde vernier 

In this type, n vernier division occupy (n + 1) main scale divi- 
sions, 

i.e. nv = (n + l)d 



-'fcr 1 ) 



v = d HJ^ (4.79) 



The least count = v - d 



= *m 



teM 



v - d = — as before (4.80) 

n 

4.93 Special forms used in vernier theodolites 

In order to provide a better break down of the graduations, the 
vernier may be extended in such a way that n vernier spaces occupy 
(mn - 1) spaces on the main scale, (m is frequently 2.) 

nv = (jnn - l)d 



The least count = md - v = md 



. d(HLzl) (4.81) 



= d 



md - v = — as before (4.82) 



4.94 Geometrical construction of the vernier scale 

In Fig. 4.74(a) the main scale and vernier zeros are coincident. 
For the direct reading vernier 10 divisions on the vernier must 
occupy 9 divisions on the main scale. Therefore 

(1) Set off a random line OR of 10 units. 

(2) Join R to V i.e. the end of the random line R to the end of 
the vernier V. 

(3) Parallel through each of the graduated lines or the random line 



INSTRUMENTAL OPTICS 



239 



to cut the main scale so that 1 division of the vernier = 0*9 divisions 
of the main scale. 



10 R 




Vernier scale 36-0 ♦ 0-3= 36-3 
reading 

Fig. 4.74 Construction of a direct reading vernier 

To construct a vernier to a given reading 

In Fig. 4.74(b) the vernier is required to read 36*3. It is thus re- 
quired to coincide at the 3rd division, i.e. 3 x 0*9 = 2*7 main scale 
division beyond the vernier index. 

Therefore coincidence will occur at (36*3 + 2*7) = 39*0 on the 
main scale and 3 on the vernier scale. 

The vernier is constructed as above in the vicinity of the point of 
coincidence. The appropriate vernier coincidence line (i.e. 3rd) is 
joined to the main scale coincidence line (i.e. 3S9*0) and lines drawn 
parallel as before will produce the appropriate position of the vernier 
on the main scale. 

In the case of the retrograde vernier, Fig. 4.75, 10 divisions on the 
vernier equals 11 divisions on the main scale, and therefore the point 
of coincidence of 3 on the vernier with the main scale value is 

36-3 - (3 x 1-1) = 36-3 - 3-3 
= 33-0 




i i i I i i i i i i i i i 

30 | 40 50 

Coincidence 
Fig. 4. 75 Construction of a retrograde vernier 



240 SURVEYING PROBLEMS AND SOLUTIONS 

Example 4.18 Show how to construct the following verniers: 

(1) To read to 10" on a limb divided to 10 minutes. 

(2) To read to 20" on a limb divided to 15 minutes. 

(3) The arc of a sextant is divided to 10 minutes. If 119 of these 
divisions are taken as the length of the vernier, into how many divi- 
sions must the vernier be divided in order to read to (a) 5 seconds (b) 
10 seconds? (ICE.) 

(1) The least count of the vernier is given by Eq. (4.77) as d/n 

10" - 10 x 60 

~ n 

n = = 60 

10 — 

Therefore the number of spaces on the vernier is 60 and the number of 
spaces on the main scale is 59. 



(2) Similarly, 20" = 11 



60 



n 

... „ = llx«! = 45 

20 — 

i.e. the number of spaces in the vernier is 45 and the number of spaces 
on the main scale is 44. 

(3) The number of divisions 119 is not required, and the calculation 
is exactly as above. 

(a) n = 10 * 60 = 120 

(b) „ , 10 *60 , 60 

10 — 

Exercises 4(b) 

3. The eccentricity of the line of collimation of a theodolite telescope 
in relation to the azimuth axis is 1/40 th of an inch. What will be the 
difference, attributable to this defect, between face right and face left 
measurement of an angle if the lengths of the drafts adjacent to the in- 
strument are 20 ft and 120 ft respectively ? 

(M.Q.B./S Ans. 17-9") 

4. A horizontal angle is to be measured having one sight elevated to 
32° 15 whilst the other is horizontal. If the vertical axis is inclined 
at 40 to the vertical, what will be the error in the recorded value ? 

(Ans. 25") 

5. In the measurement of a horizontal angle the mean angle of eleva- 
tion of the backsight is 22° 12' whilst the foresight is a depression 



INSTRUMENTAL OPTICS 241 

of 37° 10'. If the lack of vertically of the vertical axis causes the 
horizontal axis to be inclined at 50" and 40" respectively in the same 
direction, what will be the error in the recorded value of the horizontal 
angle as the mean of face left and right observations ? 

(Ans. 51") 

6. In a theodolite telescope the line of sight is not perpendicular to 
the horizontal axis in but in error by 5 minutes. In measuring a horizontal 
angle on one face, the backsight is elevated at 33° 34' whilst the fore- 
sight is horizontal. What error is recorded in the measured angle ? 

(Ans. 60") 

7. The instrument above is used for producing a level line AB by 
transiting the telescope, setting out C,, and then by changing face 
the whole operation is repeated to give C 2 . If the foresight distance 
is 100 ft, what will be the distance between the face left and face 
right positions, i.e. C,C 2 ? 

(Ans. 6-98 in.) 

8. Describe with the aid of a sketch, the function of an internal 
focussing lens in a surveyors telescope and state the advantages and 
disadvantages of internal focussing as compared to external focussing. 

In a telescope, the object glass of focal length 6 in. is located 
8 in. from the diaphragm. The focussing lens is midway between these 
when a staff 80 ft away is focussed. Determine the focal length of the 
focussing lens. 

(L.U. Ans. 4-154 in.) 

9. In testing the trunnion axis of a vernier theodolite, the instrument 
was set up at '0', 100 ft from the base of the vertical wall of a tall 
building where a well-defined point A was observed on face left at a 
vertical angle of 36° 52'. On lowering the telescope horizontally with 
the horizontal plate clamped, a mark was placed at B on the wall. On 
changing face, the whole operation was repeated and a second position 
C was fixed. 

If the distance BC measured 0*145 ft, calculate the inclination of 

the trunnion axis. , „ 

(Ans. 3' 20 ) 

10. The following readings were taken on fine sighting marks at B 
and C from a theodolite station A. 

Instrument Tq Vertical F.R.S.R. F.L.S.L. 

At Angle Vernier A Vernier B Vernier A Vernier B 

A B 72° 30 ' 292°26'30" 112°26'30" 23° 36' 24" 203° 36' 24" 

C -10° 24' 52° 39' 36" 232° 39' 36" 143°50'lo" 323° 50* 10" 
B 72° 30 ' 292°26'30" 112° 26*30" 23° 36' 24" 203° 36* 24" 

Calculate the value of the horizontal collimation error assuming 



242 SURVEYING PROBLEMS AND SOLUTIONS 

this to be the only error in the theodolite and state whether it is to the 
right or left of the line perpendicular to the trunnion axis when the in- 
strument is face left. 

Describe as briefly as possible how you would adjust the theodo- 
lite to eliminate this error. , . n _ .„ . , v 

(L.U. Ans. 8-66 left) 

11. The focal length of object glass and anallatic lens are 5 in and 
4jin. respectively. The stadia interval was 0*1 in. 

A field test with vertical staffing yielded the following: 

Inst Staff Staff Vertical Measured Horizontal 

Station Station Intercept Angle Distance (ft) 

P Q 2-30 +7° 24' 224-7 

R 6-11 -4° 42' 602-3 

Find the distance between the object glass and anallatic lens. 
How far, and in what direction, must the latter be moved so that the 
multiplying constant of the instrument is to be 100 exactly ? 

(L.U. Ans. 7-25 in.; 0*02 in. away from objective) 

12. An object is 20 ft from a convex lens of focal length 6 in. On 
the far side of this lens a concave lens of focal length 3 in. is placed. 
Their principal axes are on the line of the object, and 3 in. apart. De- 
termine the position, magnification and nature of the image formed. 
(Ans. Virtual image 43 in. away from the concave lens towards object; 

magnification 0-28) 

13. A compound lens consists of two thin lenses, one convex, the 
other concave, each of focal length 6 in. and placed 3 in apart with 
their principal axes common. Find the position of the principal focus 
of the combination when the light is incident first on (a) the convex 
lens and (b) the concave lens. 

(Ans. (a) real image 6 in. from concave lens away from object, 
(b) real image 18 in. from convex lens away from object) 

14. Construct accurately a 30 second vernier showing a reading of 

124° 23 30 ' on a main scale divided to 20 minutes. A straight line may 

be used to represent a sufficient length of the arc to a scale of 0*1 in. 

to 20 min. 

(N.R.C.T.) 

15. (a) Explain the function of a vernier. 

(b) Construct a vernier reading 4*57 in. on a main scale divided 
to 1/10 in. 

(c) A theodolite is fitted with a vernier in which 30 vernier divi- 
sions are equal to 14° 30' on a main scale divided to 30 minutes. Is the 
vernier direct or retrograde, and what is its least count ? 

(N.R.T.C. Ans. direct; 1 min.) 



INSTRUMENTAL OPTICS 243 



Bibliography 



THOMAS, W.N. , Surveying (Edward Arnold) 

SANDOVER, J. A., Plane Surveying (Edward Arnold) 

GLENDENNING, J., Principles and Use of Surveying Instruments 

(Blackie) 

CLARK, D., Plane and Geodetic Surveying, Vol. 1 (Constable) 

HIGGINS, A.L., Elementary Surveying (Longmans) 

TAYLOR, W.E., Theodolite Design and Construction (Cooke, 

Troughton & Simms) 

CURTIN, w. and LANE, R.F., Concise Practical Surveying (English 

Universities Press) 

SHEPPARD, J.S. Theodolite Errors and Adjustments (Stanley) 

The Theodolite and Its Application English version by A.H. Ward, 

F.R.I.C.S. (Wild (Heerbrugg)) 

JAMESON, A.H. Advanced Surveying (Pitman) 

higgins, A.L. Higher Surveying (Macmillan) 

SMIRNOFF, M.V., Measurements for Engineering and Other Surveys 

(Prentice -Hall) 

BANNISTER, A. and RAYMOND, s., Surveying (Pitman) 

DIBDIN, F.S.H., Essentials of Light (Macmillan) 

NELKON, M., Light and Sound (Heinemann) 

STANLEY, W.W., Introduction to Mine Surveying (Stanford University 

Press) 



LEVELLING 



5.1 Definitions 

Levelling is the process concerned with the determination of the 
differences in elevation of two or more points between each other or 
relative to some given datum. 

A Datum may be purely arbitrary but for many purposes it is taken 
as the mean sea level (M.S.L.) or Ordnance Datum (O.D.). 

A Level Surface can be defined as a plane, tangential to the 
earth's surface at any given point. The plane is assumed to be perpen- 
dicular to the direction of gravity which for most practical purposes is 
taken as the direction assumed by a plumb-bob. 

A level line, Fig. 5.1, is a line on which all points are equidistant 
from the centre of gravity. Therefore, it is curved and (assuming the 
earth to be a sphere) it is circular. For more precise determinations 
the geoidal shape of the earth must be taken into consideration. 

Horizontal line 



Vertical 




Fig. 5.1 

A Horizontal Line, Fig. 5.1, is tangential to a level line and is 
taken, neglecting refraction, as the line of collimation of a perfectly 
adjusted levelling instrument. (As the lengths of sights in levelling 
are usually less than 450 ft, level and horizontal lines are assumed to 
be the same— see §5.6 Curvature and Refraction.) 

The Line of Collimation is the imaginary line joining the inter- 
section of the main lines of the diaphragm to the optical centre of the 
object-glass. 

Mean Sea Level. This is the level datum line taken as the 



LEVELLING 



245 



reference plane. In the British Isles the Ordnance Survey originally 
accepted the derived mean sea level value for Liverpool. This has 
been superseded by a value based upon Newly n in Cornwall. 

Bench Mark (7\) (B.M.). This is a mark fixed by the Ordnance 
Survey and cut in stable constructions such as houses or walls. The 
reduced level of the horizontal bar of the mark is recorded on O.S. 
maps and plans. 

Temporary Bench Mark (T.B.M.). Any mark fixed by the observer 
for reference purposes. 

Backsight (B.S.) is the first sight taken after the setting up of the 
instrument. Initially it is usually made to some form of bench mark. 
Foresight (F.S.) is the last sight taken before moving the instrument. 
Intermediate Sight (I.S.) is any other sight taken. 

N.B. During the process of levelling the instrument and staff are 
never moved together, i.e. whilst the instrument is set the staff may 
be moved, but when the observations at one setting are completed the 
staff is held at a selected stable point and the instrument is moved 
forward. The staff station here is known as a Change Point (C.P.). 

5.2 Principles 
Let the staff readings be a, b, c etc. 



Change I.S. 
point (CR) 



Inverted staff 
reading (f) 




F.S. 



A B 


C D 
Fig. 5.2 


£ 




F 


In Fig. 5.2, 










Difference in level A to B 


= a-b 




a <b 


•'. -ve i.e. fall 


B to C 


= b-c 




b> c 


•'• +ve i.e. rise 


C to D 


= c-d 




c <d 


••• -ve i.e. fall 


DtoE 


= e-(-/) inverted staff 


•'• +ve i.e. rise 


Eto F 


= -f-g 






-ve i.e. fall 


AtoF 


= (a-fo)+ (b- 

= a - d + e - 


-c) 
g 


+ (c-d) + 


<«+/) + <-/-*) 



246 



SURVEYING PROBLEMS AND SOLUTIONS 



= (a+e)- (d+g) 
= 2B.S. -2 F.S. 
2 rises = (b-c) + (e+f) 
2 falls = (b-a)+(d-c)+(f+g) 
2 rises - 2 falls = (a + e) -(d + g) = 2B.S. - 2F.S. (5.1) 

The difference in level between the start and finish 

= 2B.S. -2F.S. = 2 rises - 2 falls (5.2) 

N.B. (1) Intermediate values have no effect on the final results, and 
thus reading errors at intermediate points are not shown up. 
(2) Where the staff is inverted, the readings are treated as nega- 
tive values and indicated in booking by a bracket or an 
asterisk. 

5.3 Booking of Readings 



5.31 Method 1, rise and fall 



B.S. 


I.S. 


F.S. 


Rise 


Fall 


Reduced Level 


a 


b 






b —a 


X 

x-(b-a) 




c 




b-c 




x- (b-a)+ (b-c) 


e 




d 




d-c 


x - (b-a)+ (b-c)- (d-c) 




[/] 




«-(-/) 




x-(b-a)+(b-c)-(d-c) 
+ (e + f) 






g 




f+g 


x-(b-a) + (b-c)- (d-c) 
+ (e+f)- (f+g) 


a + e 
- (d + g) 


d + g 


(b-c)+(e+f) 


(b-a)+ (d-c) 
+ (/+*) 





Example 5.1 Given the following readings: 

a = 2-06 e = 7'41 

b = 5-13 / = -6-84 

c = 3-28 g = 3-25 

d = 3-97 

N.B. (1) The difference between adjacent readings from the same in- 
strument position gives rise or fall according to the sign 
+ or -. 
(2) At the change point B.S. and F.S. are recorded on the same 
line. 



LEVELLING 



247 



(3) Check 2 B.S. - 2 F.S. = 2 Rise - 2 Fall before working out 
reduced levels. Difference between reduced levels at start 
and finish must equal 2 B.S. - 2 F.S. 



B.S. 


I.S. 


F.S. 


Rise 


Fall 


Reduced Level 


Remarks 


2-06 










100-00 


St. A. B.M. 100-00 A.O.D. 




5-13 






3-07 


96-93 


St. B. 




3-28 




1-85 




98-78 


St. C. 


7-41 




3-97 




0-69 


98-09 


St. D. C.P. 




[6-84] 




14-25 




112-34 


St. E. Inverted staff on girder 






3-25 
7-22 




10-09 


102-25 


St. F. 


9-47 


16-10 


13-85 


102-25 




-7-22 






-13-85 




100-00 

+ 2-25 




+ 2-25 


+ 2-25 





5.32 Method 2, height of collimation 



Height of Collimation 



B.S. 


I.S. 


F.S. 


a 


b 
c 




e 


[/] 


d 






i 


a + e 


b + c-f 


d + g 



x + a 



x+a—d+e 



Reduced Level 



x 

x + a-b 

x + a — c 

x + a-d 

x + a-d + e-(-f) 

x+a-d+e-g 



6x+5a-b-c-3d + 2e+f-g 



Arithmetical Check 

2 Height of each collimation 

x no. of applications = 3(x + a) + 2(x + a-d + e) 

= 5x + 5a - 2d + 2e 



2 Reduced levels - first 
2 I.S. 
2 F.S. 



= 5x + 5a-b-c-3d + 2e+f-g 
+ b + c - f 

+ d + g 



5x + 5a 



-2d + 2e 



Thus the full arithmetical check is given as: 

2 Reduced levels less the first + 2 I.S. + 2 F.S. should equal 
2 Height of each collimation x no. of applications (5.3) 

Using the values in Example 5.1, 



248 



SURVEYING PROBLEMS AND SOLUTIONS 



B.S. 


I.S. 


F.S. 


Height of 
Collimation 


Reduced Level 


Remarks 


2-06 






102-06 


100-00 


St. A B.M. + 100-00 A.O.D. 




5-13 






96-93 


St. B 




3-28 






98-78 


St. c 


7-41 




3-97 


105-50 


98-09 


St. D C.P. 




[6-84] 






112-34 


St. E Inverted staff on girder 






3-25 




102-25 
508-39 


St. F 


9-47 


8-41 


7-22 




7-22 


-6-84 










2-25 


1-57 





Check 508*39 + 1-57 + 7-22 



517-18 



102-06x3 = 306-18 
105-50x2 = 211-00 = 517-18 

N.B. (1) The height of collimation = the reduced level of the B.S. + 
the B.S. reading. 

(2) The reduced level of any station = height of collimation - 
reading at that station. 

(3) Whilst 2B.S.- 2 F.S. = the difference in the reduced level 
of start and finish this does not give a complete check on 
the intermediate values; an arithmetical error can be made 
without being noticed. 

(4) The full arithmetical check is needed to ensure there is no 
arithmetical error. 

On the metric system the bookings would appear thus: 

(for less accurate work the third decimal place may be omitted) 



B.S. 


I.S. 


F.S. 


Rise 


Fall 


Height of 
Collimation 


Reduced 
Level 


Remarks 


0-628 
2-259 


1-564 
1-000 

[2-085] 


1-210 
0-991 


0-564 
4-344 


0-936 
0-210 
3-076 


31-108 
32-157 


30-480 
29-544 
30-108 
29-898 
34-242 
31-166 


St. A 30-480m A.O.D. 


2-887 
2-201 


2-564 
2-085 

0-479 


2-201 


4-908 
4-222 


4-222 


154-958 




0-686 


0-686 


0-686 





Check on collimation 



154-958 + 0-479 + 2-201 = 157-638 



LEVELLING 



249 



31-108 x 3 = 93-324 
32-157 x 2 = 64-314 



157-638 



Example 5.2 Using the height of collimation calculate the respective 
levels of floor and roof at each staff station relative to the floor level 
at A which is 20 ft above an assumed datum. It is important that a 
complete arithmetical check on the results should be shown. Note that 
the staff readings enclosed by brackets thus (3*43) were taken with 
the staff reversed. 



B.S. 


I.S. 


F.S. 


Height of 
Collimation 


Reduced 
Level 


Horizontal 
Distance (ft) 


Remarks 


2-47 






22-47 


20-00 





Floor at A 




(3-43) 






25-90 





Roof at A 




3-96 






18-51 


50 


Floor at B 




(2-07) 






24-54 


50 


Roof at B 




4-17 






18-30 


100 


Floor at C 




(1-22) 






23-69 


100 


Roof at C 




3-54 






18-93 


150 


Floor at D 


(4-11) 




(2-73) 


21-09 


25-20 


150 


Roof at D 




1-96 






19-13 


200 


Floor at E 




(5-31) 






26-40 


200 


Roof at E 




2-85 






18-24 


250 


Floor at F 




(3-09) 






24-18 


250 


Roof at F 




4-58 






16-51 


300 


Floor at G 


(3-56) 




(1-16) 


18-69 


22-25 


300 


Roof at G 




2-22 






16-47 


350 


Floor at H 




(4-67) 






23-36 


350 


Roof at H 




1-15 






17-54 


400 


Floor at / 






(6-07) 




24-76 


400 


Roof at / 


+ 2-47 


+ 24-43 


-9-96 


363-91 




-7-67 


-19-79 












-5-20 


+ 4-64 




-(-9-96) 








4-76 






+ 4-76 





Checks 



1. ZB.S. -2F.S. = 4-76 = diff. in level A-I 

2. (a) 2 Reduced levels except first 363*91 
(b) 22-47 x 7 = 157-29 

21-09 x 6 = 126-54 

18-69 x 4 = 74-76 

358-59 



250 



SURVEYING PROBLEMS AND SOLUTIONS 



(c) 2F.S. + 2I.S. = -9-96 
+ 4-64 



-5-32 



5-32 



363*91 Checks with (a) 

N.B. Inverted staff readings must always be treated as negative 
values. 



Example 5.3 The following readings were taken with a level and a 
14 ft staff. Draw up a level book page and reduce the levels by 

(a) the rise and fall method, 

(b) the height of collimation method. 

2-24, 3-64, 6-03, 11-15, (12'72 and 1-48) C.P., 4'61, 6*22, 8'78, 

B.M. (102-12 ft A.O.D.), 11-41, (13'25 and 6'02) C.P., 

2-13, 5-60, 12-21. 

What error would occur in the final level if the staff had been 
wrongly extended and a plain gap of 0*04 had occurred at the 5 ft sec- 
tion joint? 

(L.U.) 



B.S. 


I.S. 


F.S. 


Rise 


Fall 


Height of 
Collimation 


Reduced 
Level 


Remarks 


2-24 


3-64 

6-03 

11-15 






1-40 
2-39 
5-12 


122-14 


119-90 
118-50 
116-11 
110-99 




1-48 


4-61 
6-22 


12-72 




1-57 
3- 13 
1-61 


110-90 


109-42 
106-29 
104-68 






8-78 






2-56 




102-12 


B.M. 102-12 A.O.D. 




11-41 






2-63 




99-49 




6-02 


2-13 
5-60 


13-25 
12-21 


3-89 


1-84 

3-47 
6-61 


103-67 


97-65 

101-54 

98-07 

91-46 




9-74 


38-18 


3-89 


32-23 


28-44 




38-18 






32-23 










28-44 


28-44 





A combined booking is shown for convenience. 
If a 0*04 ft gap occurred at the 5 ft section all readings > 5 ft will 
be 0-04 ft too small. 



LEVELLING 



251 



The final level value will only be affected by the B.S. and F.S. 
readings after the reduced level of the datum, i.e. 102*12, although the 
I.S. 8*78 would need to be treated for booking purposes as a B.S. 



i.e. 1 B.S. = 8-78+ 6*02 = 14*80 
2 F.S. - 13-25 + 12-21 = 25-46 

Difference = - 10*66 
.*. Final level = 102' 12 - 10*66 



= 91*46 



As all these values are > 5 ft no final error will be created, since 
all the readings are subjected to equal error. 

Therefore the final reduced level is correct, i.e. 91*46 ft A.O.D. 

Example 5.4 The following readings were observed with a level: 

3*75 (B.M. 112*28), 5*79, 8*42, 12*53, C.P., 4*56, 7*42, 2*18, 
1*48, C.P., 12*21, 9*47, 5*31, 2*02, T.B.M. 

(a) Reduce the levels by the Rise and Fall method. 

(b) Calculate the level of the T.B.M. if the line of collimation 
was tilted upwards at an angle of 6min and each backsight 
length was 300ft and the foresight length 100 ft. 

(c) Calculate the level of the T.B.M. if the staff was not held up- 
right but leaning backwards at 5° to the vertical in all cases. 

(L.U.) 



(a) 



B.S. 


I.S. 


F.S. 


Rise 


Fall 


Reduced 
Level 


Remarks 


3-75 


5-79 
8-42 






2-04 
2-63 


112-28 
110-24 
107-61 


B.M. 112-28 


4-56 


7-42 
2-18 


12-53 


5-24 


4-11 
2-86 


103-50 
100-64 
105-88 




12-21 


9-47 
5-31 


1-42 


0-76 
2-74 
4-16 




106-64 
109-38 
113-54 








2-02 


3-29 




116-83 


T.B.M. 


20-52 


15-97 


16-19 


11-64 




15-97 






11-64 








4-55 






4-55 




4-55 





252 



SURVEYING PROBLEMS AND SOLUTIONS 



39 




300 ft 



100ft 



Fig. 5.3 

(b) In Fig. 5.3, 

True difference in level = (a-3e) - (b-e) = (a-b) - 2e 
where e = 100 tan 06' = 100 x 06' radian8 

= 1°° x 6 x 60 = 0-175 per 100 ft 
206 265 ~ 

Total length of backsight = 3 x 300 = 900 ft 

of foresight = 3 x 100 = 300 ft 

Effective difference in length = 600 ft 

.". Error = 6 x 0*175 = 1*050 ft 

i.e. B.S. readings are effectively too large by 1*05 ft. 

/. True difference in level = 4*55 - 1*05 = 3*50 ft 

.'. Level of T.B.M. = 112*28 + 3*50 = 115*78 ft A.O.D. 

(c) If the staff was not held vertical the readings would be too large, 
the value depending on the staff reading. 




True reading = observed reading x cos 5° 
Line of sight Apparent difference 

in level 2 B.S. - 2F.S. = 4*55 
True difference 

in level = 2 B.S. cos 5° - 2 F.S. cos 5° 
= (2 B.S. -2 F.S.) cos 5° 



Fig. 5.4 



= 4*55 cos 5° = 4*53 
Level of T.B.M. = 112*28 + 4*53 = 116*81 ft 



Example 5.5 Missing values in booking. It has been found necessary 
to consult the notes of a dumpy levelling carried out some years ago. 



LEVELLING 



253 



Whilst various staff readings, rises and falls and reduced levels are 
undecipherable, sufficient data remain from which all the missing 
values can be calculated. 



















B.S. 


I.S. 


F.S. 


Rise 


Fall 


Level A.O.D. 


Distance 


Remarks 


2.36 








1-94 


121.36 



100 


B.M. on Watson House 


4-05 


4-31 
6.93 


7-29 




4.46 


113-32 


200 
300 
400 
500 
600 








7-79 




0-63 




715 


Peg 36 




3-22 




1-58 




112-01 


800 
900 






6-53 


5-86 






113-53 


1000 
1100 
1200 












3-10 




1286 


B.M. on boundary wall 


14-96 






21-18 







Calculate the missing values and show the conventional arithmet- 
ical checks on your results. 













Reduced 






B.S. 


I.S. 


F.S. 


Rise 


Fall 


Level A.O.D. 


Distance 


Remarks 


2-36 


(a) 








121-36 





B.M. on Watson House 




4-30 






1-94 
(b) 


119-42 


100 




4-05 


(c) 


7-29 




2-99 


116-43 


200 






8-51 




(d) 


4-46 


111-97 


300 






4-31 




4-20 


(e) 


116-17 


400 






6-93 






2-62 


113-55 


500 






CO 
















7-16 






0-23 


113-32 


600 




(g) 
















4-80 




7-79 




0-63 


112-69 


715 


Peg 36 




3-22 




1-58 




114-27 


800 






(h) 
















5-48 






2-26 
0) 


112-01 


900 






6-53 






1-05 


110-96 


1000 






(k) 
















3-96 




2-57 




113-53 


1100 




(m) 








(1) 








3-75 




5-86 
(n) 




1-90 


111-63 


1200 








6-85 




3-10 


108-53 


1286 


B.M. on boundary wall 


14-96 


27-79 


8-35 


21-18 






27-79 








8-35 


-12-83 






-12-83 


-12-83 





254 



SURVEYING PROBLEMS AND SOLUTIONS 



Notes: 

(a) 4-30 I.S. is deduced from fall 1-94. 

(b) Fall 2-99 is obtained from I.S.-F.S., 4-30-7-29. 

(c) 8-51 as (a). 

(d) Rise 4-20 as (b). 

(e) Fall 2-62 as (b). 

(f ) Reduced levels 113-32 - 113-55 gives fall 0-23, which gives staff reading 
7-16. 

(g) 4-80 B.S. must occur on line of F.S. —deduced value from rise 1-58 with 
I.S. 3-22. 

(h) 5-48 as (f). 

(j) 1-05 as normal. 

(k) 3-96 as (f). 

(1) Fall 1-90 normal. 

(m) 3-75 B.S. must occur opposite 5-86 F.S. Value from 2b.S. 14-96. 

(n) 6-85 from fall 3-10. 

Checks as usual. 

Exercises 5(a) (Booking) 

1. The undernoted staff readings were taken successively with a 
level along an underground roadway. 



Staff Readings 


Distances from A 


Remarks 


5-77 





B.S. to A 


2-83 


120 


I.S. 


0-30 


240 


F.S. 


5-54 




B.S. 


1-41 


360 


I.S. 


3-01 


480 


F.S. 


2-23 




B.S. 


2-20 


600 


I.S. 


1-62 


720 


F.S. 


7-36 




B.S. 


5-52 


840 


I.S. 


0-71 


960 


F.S. 


4-99 




B.S. 


2-25 


1080 


F.S. to B 



Using the Height of Collimation method, calculate the reduced 
level of each staff station relative to the level of A, which is 
6010*37 ft above an assumed datum of 10000ft below O.D. 

Thereafter check your results by application of the appropriate 
method used for verifying levelling calculations derived from heights 
of collimation. 

(M.Q.B./S Ans. Reduced level of B 6028*37) 



LEVELLING 255 

2. A section levelling is made from the bottom of a staple pit A to 
the bottom of a staple pit B. Each group of the following staff read- 
ings relates to a setting of the levelling instrument and the appropriate 
distances from the staff points are given. 
From bottom of staple pit A. 



Staff Readings 


Distance (ft) 


4-63 




2-41 


72 


0-50 


51 


7-23 




4-80 


32 


3-08 


26 


1-02 


45 



4-09 




3-22 


48 


1-98 


53 


1-47 




3-85 


52 


6-98 


46 


1-17 




2-55 


106 


2-16 




3-64 


54 


5-27 


45 



To bottom of staple pit B. 

The top of staple pit A is 8440ft above the assumed datum of 
10000ft below O.D. and the shaft is 225ft deep. 

(a) Enter the staff readings and distances in level book form, com- 
plete the reduced levels and apply the usual checks. 

(b) Plot a section on a scale 100 ft to lin. for horizontals and 10 ft 
to lin. for verticals. 

(c) If the staple pit B is 187*4 ft deep what is the reduced level at 
the top of the shaft? 

(M.Q.B./UM Ans. 8404-85 ft) 

3. Reduce the following notes of a levelling made along a railway 
affected by subsidence. Points A and B are outside the affected area, 
and the grade was originally constant between them. 

Find the original grade of the track, the amount of subsidence at 
each chain length and the maximum grade in any chain length. 



256 



SURVEYING PROBLEMS AND SOLUTIONS 



B.S. 


I.S. 


F.S. 


Distance (links) 


Remarks 


10-15 


8-84 





100 


A 


7-58 


6-65 


7-69 


200 
300 




4-21 


2-72 


5-50 


400 
500 




8-21 


3-76 
2-38 


5-55 


600 
700 


Not on track 






1-09 


800 


B 



(M.Q.B./M Ans. Grade 1*29 in 100 links; 
subsidence +0*02, -0*12, -0'48, 
-0-62, -0-42, -0-90 
max. grade l'62ft in 100 links be- 
tween 500 - 600 ft) 

4. The following staff readings were observed in the given order 
when levelling up a hillside from a temporary bench mark 135* 20 ft 
A.O.D. With the exception of the staff position immediately after the 
bench mark, each staff position was higher than the preceding one. 

Enter the readings in level book form by both the rise and fall and 
collimation systems. These may be combined, if desired, into a single 
form to save copying. 

4*62, 8-95, 6-09, 3'19, 12*43, 9-01, 5*24, 1*33, 10*76, 6*60, 

2-05, 13-57, 8-74, 3*26, 12'80, 6*33, 11-41, 4*37. 

(L.U.) 
N.B. There are alternative solutions 

5. The undernoted readings, in feet, on a levelling staff were taken 
along a roadway AB with a dumpy level, the staff being held in the 
first case at a starting point A and then at 50ft intervals: 2*51, 3*49, 
[2-02], 6-02, 5-00. 

The level was then moved forward to another position and further 
readings taken. These were as follows, the last reading being at B: 
7-73, 4-52, [6-77], 2'22, 6*65. 

The level of A is 137'2ft A.O.D. 

Set out the staff readings and complete the bookings. 

Calculate the gradient from A to B. 

(Figures in brackets denote inverted staff readings.) 

(R.I.C.S./M Ans. 1 in 284) 

6. An extract from a level book is given below, in which various 
bookings are missing. Fill in the missing bookings and re-book and 



LEVELLING 



257 



complete the figures by the Height of Collimation method. 



B.S. 


I.S. 


F.S. 


Rise 


Fall 


Reduced Level 












154-86 




7-62 






3-70 






2-32 




5-30 






7-11 










147-72 






5-55 


1-56 




149-28 




7-37 




2-00 




151-28 




8-72 






1-61 


149-93 






4-24 


6-09 




154-41 



All the figures are assumed correct. 

(I.C.E.) 

7. The following figures are the staff readings taken in order on a 
particular scheme, the backsights being shown in italics: 

2'67, 7*12, 9-54, 8'63, 10-28, 12'31, 10*75, 6*23, 7*84, 
9-22, 5-06, 4-18, 2' 11. 

The first reading was taken on a bench mark 129*80 O.D. 
Enter the readings in level book form, check the entries and find 
the reduced level of the last point. 

Comment on your completed reduction. 

(L.U./E) 

5.4 Field Testing of the Level 

Methods available are (1) by reciprocal levelling, (2) the two-peg 
methods. 



5.41 Reciprocal levelling method 

In Fig. 5.5, the instrument is first set at A of height a. The line 
of sight is assumed to be inclined at an angle of elevation + a giving 
an error e in the length AB. The reading on the staff at B is b. 

Difference in level A - B = a - (b-e) (5.4) 



Fig. 5.5 




Difference in level 



258 



SURVEYING PROBLEMS AND SOLUTIONS 




Fig. 5.6 

In Fig. 5.6, the instrument is set at B of height ft, and the reading 
on the staff at A is a r 

Difference in level A - B = (a,-e) - b, (5.5) 

Thus, from Eqs (5.4) and (5.5), 

Difference in level = a - b + e (5.4) 

= a, - fe, - e (5.5) 

Adding (5.4) and (5.5), = i[(a-b) + (a,-ft,)] (5.6) 



By subtracting Eqs. (5.4) and (5.5), the error in collimation 

e = i[(a,-5,)-(a-5)] 



(5.7) 



Example 5.6 A dumpy level is set up with the eyepiece vertically 
over a peg A. The height from the top of A to the centre of the eye- 
piece is measured and found to be 4*62 ft. A level staff is then held on 
a distant peg B and read. This reading is 2*12 ft. The level is then 
set over B. The height of the eyepiece above B is 4* 47 ft and a read- 
ing on A is 6* 59 ft. 

(1) What is the difference in level between A and B ? 

(2) Is the collimation of the telescope in adjustment? 

(3) If out of adjustment, can the collimation be corrected without 
moving the level from its position at 6? (I.C.E.) 

(1) From Eq. (5.6), 

Difference in level (A-B) = \ [(4- 62 -2-1 2) + (6* 59 -4* 47)] 

= \ [2-50 + 2-12] 

= +2-31 ft 

(2) From Eq. (5.7), 

Error in collimation e = | [2*12 - 2*50] 

= - 0-19 ft per length AB. 
(i.e. the line of sight is depressed.) 



LEVELLING 259 

(3.) True staff reading at A (instrument at B) should be 

6-59 -(-0-19) 
= 6-59 + 0*19 = 6-78 ft. 
The cross-hairs must be adjusted to provide this reading. 

5.42 Two-peg method 

In the following field tests the true difference in level is ensured 
by making backsight and foresight of equal length. 



/>-• 




a-b 



Fig. 5.7 
Assuming the line of collimation is elevated by a, 
the displacement vertically = dtana 
Thus, if B.S. = F.S., dtana = e in each case. 

•*• True difference in level AB = (a-e)-(b-e) 

= a ~ b 
Method (a). Pegs are inserted at A and B so that the staff reading 
a = b when the instrument is midway between A and B. The instru- 
ment may now be moved to A or B. 



a,- 2»=^ 




Fig. 5.8 

In Fig. 5.8, if the height of the instrument at B is fc, above peg 
B, the staff reading at peg A should be b, if there is no error, i.e. if 
a= 0. 

If the reading is a, and the distance AB = 2d, then the true read- 



260 



SURVEYING PROBLEMS AND SOLUTIONS 



ing at A should be 



a, - 2e 



:. e = |[a,-b,] 



(5.8) 



Method (b). The instrument is placed midway between staff positions 
A and B, Fig. 5.9. Readings taken give the true difference in level = 
a-b. 




a,-kt 



Fig. 5.9 

The instrument is now placed at X so that 

XA = kXB 

where k is the multiplying factor depending on the ratio of AX/BX. 
If AX = kBX, then the error at A = ke 

True difference in level = (a y -ke) - (fe, -e) = a-b 

a,-6,-(&-l)e = a-b 

(a,-b,)-(a-fc) 



Error per length BX 



e = 



* ~ x (5.9) 
N.B. (1) If the instrument is placed nearer to A than B, k will be 
less than 1 and /c-1 will be negative (see Example 5.8). 
(2) If the instrument is placed at station B, then Eq. (5.9) is 
modified as follows: 

(a,-£) - b, = a-b 

.'. E = (a,-b,)-(a-b) (5.10) 

where E is the error in the length AB (see Example 5.9). 

Example 5.7 (a) When checking a dumpy level, the following readings 
were obtained in the two-peg test: 

Level set up midway between two staff stations A and B 400 ft 
apart; staff readings on A 5'75ft and on B 4*31 ft. 

Level set up 40 ft behind B and in line AB; staff reading on B 
3-41 ft and on A 4*95 ft. 



LEVELLING 261 

Complete the calculation and state the amount of instrumental 
error. 

(b) Describe the necessary adjustments to the following types of 
level, making use of the above readings in each case: 

(i) Dumpy level fitted with diaphragm screws and level tube 

screws . 
(ii) A level fitted with level screws and tilting screw. 



(a) By Eq. (5.9) 

e 



Check 



(4-95 - 3-41) - (5-75 - 4-31) 

11 - 1 
1-54 - 1-44 



10 



= 0*01 per 40 ft 



With instrument 40 ft beyond B, 

Staff reading at A should be 4*95 -(11 x 0-01) 

= 4-95- 0-11 = 4- 84 ft 
Staff reading at B should be 3'41 - 0*01 = 3'40ft 
Difference in level = 1*44 
This agrees with the first readings 5*75 - 4'31 = 1'44 

(b) (i) With a dumpy level the main spirit level should be first ad- 
justed. 

The collimation error is then adjusted by means of the dia- 
phragm screws until the staff reading at A from the second 
setting is 4*84 and this should check with the staff reading 
at B of 3*40. 

(ii) With a tilting level, the circular (pill-box) level should be 
first adjusted. 

The line of sight should be set by the tilting screw until the 
calculated readings above are obtained. 
The main bubble will now be off centre and must be central- 
ised by the level tube screws. 

Example 5.8 (a) Describe with the aid of a diagram the basic prin- 
ciples of a tilting level, and state the advantages and disadvantages 
of this type of level compared with the dumpy level. 

(b) The following readings were obtained with a tilting 
level to two staves A and B 200ft apart. 



262 



SURVEYING PROBLEMS AND SOLUTIONS 



Position of Instrument 

Midway between A and B 

10 ft from A and 200 ft from B 



Reading at 
A (ft) 

5-43 
6-17 



Reading at 
B(ft) 

6-12 
6-67 



What is the error in the line of sight per 100 ft of distance and how 
would you adjust the instrument? (R.I.C.S. L/Inter.) 

Part (b) illustrates the testing of a level involving a negative 

angle of inclination of the line of collimation and a frac- 
tional k value. 

Using Eq. (5.9), 

10 



e - ("!-».>-("-»> where 
k- 1 

(6-17-6-67)- (5-43 -6- 12) 



BX 200 



1/20 - 1 
20(- 0-50 +0-69) 



19 



= - 0-01 x 20 = - 0-2 ft per length BX 



Check 



i.e. e = - 0-2 per 200ft = - Q-l per 100ft 

At X, Reading on A should be 6*17 + 0*01 = 6*18 

Reading on B should be 6*67 + 0-20 = 6*87 

difference in level = - 0*69 



Alternative solution from first principles (Fig.5.10) 




6-67 -209 



Fig. 5.10 
True difference in level = (6'17-e) - (6'67-20e) = (5'43-6'12) 

i.e. 19e- 0-50 = -0*69 
0-19 



e = 



19 

= -0*01 per 10 ft 
= -0*1 ft per 100 ft 



LEVELLING 



263 



Example 5.9 The following readings taken on to two stations A and 
B were obtained during a field test of a dumpy level. Suggest what 
type of error exists in the level and give the magnitude of the error as 
a percentage. How would you correct it in the field? 



B.S. 
6-21 

4-99 



By Eq. (5.10) 



F.S. 



5-46 



4-30 



Remarks 

Staff at station A A-B 200ft apart 

Staff at station B Instrument midway 

Staff at station A AB 200 ft apart 

Staff at station B Instrument v. near to B 



(R.I.C.S./G) 



E = („,-&,) -(a-*) 

= (4-99-4-30) -(6-21-5-46) 

= 0-69-0-75 

= - 0-06 ft per 200 ft 

= -0-03 ft per 100 ft. 



Check 



4-99 + 0-06 = 
4-30 + 

True difference in level = 



5-05 
4-30 
0-75 ft 
6-21 - 5-46 



= 0-75 ft. 



Example 5.10 In levelling up a hillside to establish a T.B.M. (tem- 
porary bench mark) the average lengths of ten backsights and ten 
foresights were 80ft and 40ft respectively. 

As the reduced level of the T.B.M. of 82*50ft A.O.D. was in doubt, 
the level was set up midway between two pegs A and B 200 ft apart, 
the reading on A being 4*56 and that on B 5*24. When the instrument 
was moved 40ft beyond B on the line AB produced, the reading on A 
was 5*34 and on B 5*88. 

Calculate the true value of the reduced level. 

True difference in level A-B = 4-56-5*24 = -0*68 

When set up 40 ft beyond B, 

(5'34-6e)-(5-88-e) = -0-68 
5-34 - 5-88 - Se = - 0*68 



Se = 



e = 



0-14 

0-028 ft/40 ft. 



264 SURVEYING PROBLEMS AND SOLUTIONS 



Check 




True readings should have been: 




at A 5-34-0-168 = 


5-172 


at B 5-88 - 0-028 = 


5-852 




- 0-680 



Error in levelling = 0*028 ft per 40 ft 

Difference in length between B.S. and F.S. per set = 80-40 = 40 ft 

per 10 sets = 400 ft 
.*• Error = 10 x 0*028 = - 0*28 ft 
.". True value of T.B.M. = 82'50 - 0'28 

= 82-22 ft A.O.D. 



Exercises 5(b) (Adjustment) 

8. Describe how you would adjust a level fitted with tribrach screws, 
a graduated tilting screw and bubble-tube screws, introducing into 
your answer the following readings which were taken in a 2 peg test: 

Staff stations at A and B 400ft apart. 

Level set up halfway between A and B : staff readings on A 
4-21 ft, on B 2-82 ft. 

Level set up 40 ft behind B in line AB : staff readings on A 
5-29 ft, on B 4*00 ft. 

Complete the calculation and show how the result would be used 
to adjust the level. 

(L.U. Ans. Error -0*01 per 40ft) 

9. A modern dumpy level was set up at a position equidistant from 
two pegs A and B. The bubble was adjusted to its central position 
for each reading, as it did not remain quite central when the telescope 
was moved from A to B. The readings on A and B were 4*86 and 
5*22 ft respectively. The instrument was then moved to D, so that the 
distance DB was about five times the distance DA, and the readings 
with the bubble central were 5*12 and 5*43 ft respectively. Was the 
instrument in adjustment? 

(I.C.E. Ans. Error = 0*0125 ft at A from D) 

10. The table gives a summary of the readings taken when running a 
line of levels A, B, C, D. The level used was fitted with stadia hairs 
and had tacheome trie constants of 100 and 0. For all the readings the 
staff was held vertically. 



LEVELLING 



265 



Reduce the levels shown in the table 



Position of 




Backsight 






Foresight 




Staff 


Top 


Middle 


Bottom 


Top 


Middle 


Bottom 


A 


6-22 


4-37 


2-52 








B 


4-70 


2-94 


1-18 


11-06 


9-38 


7-70 


C 


7-63 


5-27 


2-91 


9-32 


7-43 


5-54 


D 








8-17 


6-04 


3-91 



It was suspected that the instrument was out of adjustment and to 
check this, the following staff readings were taken, using the centre 
hair of the level diaphragm; P and Q are 300 ft apart. 



Instrument Station 


Staff at P 


Staff at Q 


Near P 
Near Q 


4-65 
2-97 


8-29 
6-17 



Find the true reduced level of D if the reduced level of A is 
125-67 ft A.O.D. 

(Ans. 115-36 ft A.O.D.) 

11. A level was set up on the line of two pegs A and B and 
readings were taken to a staff with the bubble central. If A and B 
were 150 metres apart, and the readings were 2*763 m and 1*792 m re- 
spectively, compute the collimation error. The reduced levels were 
known to be 27*002 m and 27*995 m respectively. 

The level was subsequently used, without adjustment, to level 
between two points X and Y situated 1 km apart. The average length 
of the backsights was 45 m and of the foresights 55 m. What is the error 
in the difference in level between X and Y? 

(N.R.C.T. Ans. 30*5" depressed; error + 0*0145m) 



12. The following readings were taken during a 'two-peg' test on a 
level fitted with stadia, and reading on a vertical staff, the bubble 
being brought to the centre of its run before each reading. 

The points A, B, X and Y were in a straight line, X being mid- 
way between A and B and Y being on the side of B remote from A. 

(A) If the reduced level of A is 106*23, find the reduced level 
of B. 

(B) Explain what is wrong with the instrument and how you would 
correct it. 

(C) Find what the centre hair readings would have been if the 
instrument had not been out of adjustment. 



266 



SURVEYING PROBLEMS AND SOLUTIONS 



Instrument at 


Staff at 


Staff Readings 


X 


A 


5-56 
4-81 
4-06 


X 


B 


8-19 
7-44 
6-69 


Y 


A 


5-32 
3-72 
2-12 


Y 


B 


6-31 
6-21 




6-11 


(R.I.C.S./I 


VM. Ans. 


103-60; 4-74, 7-37, 3*57, 6*20) 



13. A level set up in a position 100 ft from peg A and 200 ft from peg 
B reads 6*28 on a staff held on A and 7*34 on a staff held on B, the 
bubble having been carefully brought to the centre of its run before 
each reading. It is known that the reduced levels of the tops of the 
pegs A and B are 287*32 and 286*35 ft O.D. respectively. 

Find (a) the collimation error and 

(b) the readings that would have been obtained had there 
been no collimation error. 

(L.U. Ans. (a) 0*09ft per 100ft, (b) 6*19; 7* 16ft) 

14. P and Q are two points on opposite banks of a river about 100 yds 
wide. A level with an anallatic telescope with a constant 100 is set up 
at A on the line QP produced, then at B on the line PQ produced, 
and the following readings taken on to a graduated staff held vertically 
at P and Q. 



From 



To 



Staff Readings in ft 
Upper Stadia Collimation Lower Stadia 



P 
Q 
P 

Q 



5-14 
3-27 

10-63 
5-26 



4-67 


4-20 


1-21 


below ground 


8-51 


6-39 


4-73 


4-20 



What is the true difference in level between P and Q and what is 
the collimation error of the level expressed in seconds of arc, there 
being 206 265 seconds in a radian? 

(I.C.E. Ans. 3*62ft; 104" above horizon) 



LEVELLING 



267 



5.5 Sensitivity of the Bubble Tube 

The sensitivity of the bubble tube depends on the radius of curva- 
ture (R) and is usually expressed as an angle (0) per unit division 
(d) of the bubble scale. 

5.51 Field test 

Staff readings may be recorded as the position of the bubble is 
changed by a footscrew or tilting screw. Readings at the eye and ob- 
jective ends of the bubble may be recorded or alternatively the bubble 
may be set to the exact scale division. 




Reading (a) 



sac a— b 



Reading (M 











Fig. 


5.11 




In 


Fig. 


5.11, 


tan(n0) = 


I 


but 


e 


is very small 


(usually 1 to 60 seconds) 










•'• "0rad = 


S 
I 










#rad = 


s 
nl 










d— = 


206265 s 



(5.11) 



nl 



(5.12) 
(5.13) 



where s = difference in staff readings a and b 

n = number of divisions the bubble is displaced between readings 
/ = distance from staff to instrument. 
If d = length of 1 division on the bubble tube, then 

d = R.d TBd 



R = * 



ndl 
s 



(5.14) 
(5.15) 



268 SURVEYING PROBLEMS AND SOLUTIONS 

5.52 - E correction 

If the bubble tube is graduated from the centre then an accurate 
reading is possible, particularly when seen through a prismatic reader, 
Fig. 5.12. 



Objective end Eye end 

20 16 12 8 4 4 8 12 16 20 



-+I-H — I — I — I — I — h — I — I — I — I — lr—f- 



I I I l i 






fl 



T 






Fig. 5.12 



If the readings at the objective end are 0, and Z and those at 
the eye end E, and E 2 , then the movement of the bubble in n divi- 
sions will equal 



or 



(0 1 -E 1 ) + (0 2 -£ 2 ) 

2 
(0,+0 2 )-(E, + E 2 ) 



(5.16) 



The length of the bubble will be 



+ E 



(5.17) 



— F 
The displacement of the bubble will be — - — (5.18) 



If > E the telescope is elevated, 
< E the telescope is depressed. 

5.53 Bubble scale correction 

With a geodetic level, the bubble is generally very sensitive, say 
1 division = 1 second. 

Instead of attempting to line up the prismatically viewed ends of 



LEVELLING 269 

the bubble, their relative positions are read on the scale provided and 
observed in the eyepiece at the time of the staff reading. 

The correction to the middle levelling hair is thus required. 

ByEq. (5.13), 

R 206 265 s 

''sec i 

nl 

Transposing gives e = "*£.,. (5.19) 

206 265 

where e - the error in the staff reading 

6 = the sensitivity of the bubble tube in seconds 

n = the number of divisions displaced 

/ = length of sight. 

Example 5.11 Find the radius of curvature of the bubble tube attached 
to a level and the angular value of each 2 mm division from the follow- 
ing readings taken to a staff 200 ft from the instrument. 
(2 mm = 0-006 56 ft). 

Staff Readings 3-510 3-742 

Bubble Readings Eye End 18-3 6-4 

Objective End 3-4 15-3 

By Eq. (5.16), n = | [(18*3 - 3-4) + (15*3 - 6-4)] 

- |[14-9 + 8-9] 

= 11*9 divisions. 

By Eq. (5.13), 6 = 206265 s 

nl 

206 265 x (3-742-3-510) 
11-9 x 200 
= 20 sec 



By Eq. (5.15), R = 



ndl 
s 

0-006 56 x 200 x 11-9 



0-232 
= 67- 3 ft 
In the metric system the above readings would be given as: 
Staff readings l'070m M41m 

Distance between staff and level = 60- 96 m 



270 SURVEYING PROBLEMS AND SOLUTIONS 

n*. L r, ,r,^ „ 206 265 x (1-141 -1-070) „. 
Then, by Eq. (5.13), - IT9TW96 = ^^ 

by Eq. (5.15), R . °-° 02 x^ 6 * "' 9 . 2H3« 

Example 5. 12 The following readings were taken through the eye- 
piece during precise levelling. What should be the true middle hair 
reading of the bubble value if 1 division is 1 second. The stadia con- 
stant of the level is x 100. 



Stadia Readings 
Top Middle Bottom 



6-3716 5-507 4 4-643 1 
By Eq. (5.18), n = ° " 



Bubble Scale Readings 
E O 



10-6 8*48 



Then by Eq. (5.19), e = 



2 




8-4 - 10-6 
2 




- 1-1 




nW 
206 265 




- 1-1 x 100(6-3716- 


- 4-643 1) x 1" 


206 265 




110 x 1-7285 





206265 
= - 0*0009 
.*. True middle reading should be 5' 507 4 + 0*0009 

= 5-508 3 

Exercises 5(c) (Sensitivity) 

15. A level is set with the telescope perpendicular to two footscrews 
at a distance of 100 ft from a staff. 

The graduations on the bubble were found to be 0*1 in. apart and 
after moving the bubble through 3 divisions the staff readings differed 
by 0-029 ft. 

Find the sensitivity of the spirit bubble tube and its radius of 
curvature. 

(Ans. e ^ 20 seconds; R = 86*2 ft) 

16. State what is meant by the term 'sensitivity' when applied to a 
spirit level, and discuss briefly the factors which influence the choice 



LEVELLING 



271 



of spirit level of sensitivity appropriate for the levelling instrument of 
specified precision. 

The spirit level attached to a levelling instrument contains a 
bubble which moves 1/10 in. per 20 seconds change in the inclination 
of the axis of the spirit level tube. Calculate the radius of curvature 
of the spirit level tube. 

(M.Q.B./S Ans. 85*94 ft) 



5.54 Gradient screws (tilting mechanism) 

On some instruments the tilting screw is graduated as shown in 
Fig. 5.13. 

The vertical scale indicates 
the number of complete revolutions 
whilst the horizontal scale indi- 
cates the fraction of a revolution. 

The positive and negative tilt 10- 

of the telescope are usually shown 
in black and red respectively and 
these must be correlated with simi- 
lar colours on the horizontal scale. 

The gradient of the line of sight 
is given as 1 in x. 

where 1/x = nr (5.20) 
where n = numbers of revs. 

r = the ratio of 1 rev (fre- 
quently 1/1000). Fig. 5.13 



mrrrrrm 
fio 40 



-5 



Vertical 
scale 



30 20 v\ 



50 60 70 80 90 



Horizontal scale 



Using the gradient screw, it is also possible to obtain the approx- 
imate distance by taking staff readings. 



(5.21) 



If gradient = 


j- = nr 


then 


L = £ 
nr 


where s = staff intercept 




L = length of sight 




Example 5.13 Staff reading (a) = 


6-32 


00 = 


6-84 


Number of revs (n) = 


6-35 


Gradient ratio (r) = 


1/1000 



272 



SURVEYING PROBLEMS AND SOLUTIONS 

(6-84 - 6-32) x 1 000 



Then L = 



6-35 



520 



6-35 
= 81- 88 ft 

5.6 The Effect of the Earth's Curvature and Atmospheric Refraction 

5.61 The earth's curvature 

Over long distances the effect of the earth's curvature becomes 
significant. 

Let the error due to the earth's curvature E = AC. 



■g Refracted 
line of sight 




In Fig. 5.14, 




ACAD = 


TA 2 


.*. AC = 


TA 2 
AD 




L 2 




2R + AC 


i.e. E ~ 


L 2 
2R 



Fig. 5.14 
(intersecting chord and tangent) 

(where L = length of sight ~ TA) 

(as E is small compared with R) (5.22) 



Alternatively, by Pythagoras, 

AO 2 - 0T 2 +AT 
i.e. (E + Rf = R 2 +L 2 



E 2 + 2RE + /T 
E 



R 2 +L' 



2R + E 



~ — as above. 
2R 



LEVELLING 273 

As R, the radius of the earth, is ^ 3 960 miles (^6 370 km). 
E ~ (5 280L) 2 

2 x 5280 x 3960 

5280L 2 
7920 

= 0-667 L 2 ft. (5.23) 

where L = length of sight in miles; 
or metric values give E = 0'0785 L 2 metres (where L = length in km) 

5.62 Atmospheric refraction 

Due to variation in the density of the earth's atmosphere, affected 
by atmospheric pressure and temperature, a horizontal ray of light TA 
is refracted to give the bent line TB. 

If the coefficient of refraction m is defined as the multiplying 
factor applied to the angle TO A (subtended at the centre) to give the 
angle ATB, ^ 

Angle of refraction A TB = m TO A 

= 2m ATC 

As the angles are small 

AB : AC : : ATB : ATC 

Then AB = 2mA J C x AC 
ATC 

= 2m AC (5.24) 

The value of m varies with time, geographical position, atmos- 
pheric pressure and temperature. A mean value is frequently taken as 
0-07. 

/. AB = 0-14 AC 

Error due to refraction ^ ±AC (5.25) 

~ 0M567_L 2 = . 095L a (526 ) 

5.63 The combined effect of curvature and refraction 

The net effect e = BC 

= AC - AB 
I 2 L z 

2R 2R 

= Ll[l-2m] (5.27) 

ZK 



274 SURVEYING PROBLEMS AND SOLUTIONS 

If m is taken as 0*07 

then e = 0-667 L 2 (1 - 0- 14) 
= 0-667 L 2 x 0-86 
= 0-574 L 2 ft (5.28) 

or metric value e m = 0*0673 L metres 
Alternatively, taking refraction as 1/7 of the curvature error, 



6 

7 

= 0-572 L* - 0-57 V 



e = | x 0-667 L 



Example 5.14 Effect of curvature 

1. What difference will exist between horizontal and level lines at 
the following distances? 









(a) 1 mile 


(d) 100 miles 








(b) 220 yards 


(e) 1 km 








(c) 5 miles 


(f) 160 km 


(a) 


E a 


= 


0-667 L 2 ft (Lin miles) 






= 


0-667 ft 


8-004 inches 


(b) 


E b 


oc 


L 2 




thus 


E b 


= 


0-667 (|) 2 - 


0*667 _ 0-010 4 ft 
64 


(c) 


E c 


= 


0-667 x 5 2 


25 x 0-667 

66 ' 7 - 16-675 ft 
4 


(d) 


E d 


= 


0-667 x 100 2 = 


0-667 x 1000 
6670 ft 


(e) 


Ee 


= 


0-0785 x l 2 = 


0-078 5 m 


(0 


E f 


= 


0-078 5 x 160 2 = 


2 009-6 m 



In ordinary precise levelling it is essential that the lengths of the 
backsight and foresight be equal to eliminate instrumental error. This 
is also required to counteract the error due to curvature and refraction, 
as this error should be the same in both directions providing the clim- 
atic conditions remain constant. To minimise the effect of climatic 
change the length of sights should be kept below 150 ft. 

In precise surveys, where the length of- sight is greater than this 
value and climatic change is possible, e.g. crossing a river or ravine, 
'reciprocal levelling' is employed. 



LEVELLING 



275 



Exercises 5(d) (Curvature and refraction) 

17. Derive the expression for the combined curvature and refraction 
correction used in levelling practice. 

If the sensitivity of the bubble tube of a level is 20 seconds of 
arc per division, at what distance does the combined curvature and re- 
fraction correction become numerically equal to the error induced by 
dislevelment of one division of the level tube. 

(R.I.C.S. D/M Ans. 4 742*4 ft) 

18. A geodetic levelling instrument which is known to be in adjust- 
ment is used to obtain the difference in level between two stations A 
and B which are 2430ft apart. The instrument is set 20ft from B on 
the line AB produced 

If A is 1*290 ft above B, what should be the reading on the staff 
at A if the reading on the staff at B is 4*055 ft. 

(M.Q.B./S Ans. 2*886ft) 

5.64 Intervisibility 

The earth's curvature and the effect of atmospheric refraction 
affect the maximum length of sight, Fig. 5.15. 



ft, = 0*57 d z 

ft, = 0*57 (D - df 




Fig. 5.15 

This will give the minimum height ft 2 at C which can be observed 
from A height ft, assuming the ray grazes the surface of the earth or 
sea. 

With intervening ground at B 
In Fig. 5.16, 

let the height of AA^ = ft, 



of BB 3 = h 2 = SB, + S,B 2 + B 2 B 3 
= 5280dtana + ft, + 0*57 d' 
of CC 3 = h 3 



= 5280 D tan £ + ft, + 0*57 D' 
If C is to be visible from A then a ^ /8. 
If a = /3, then 



tana = 



h 2 - /i, - 0*57d' 
5280d 



ft, - 0-57P' 
5280D 



(5.29) 
(5.30) 

(5.31) 



276 



SURVEYING PROBLEMS AND SOLUTIONS 




Fig. 5.16 



Thus the minimum height 



d 



§-0-57.) 



K = jj(K -h,- 0-57 D 2 ) + h, + 0-57d ; 

- £*(D-d) 

Clendinning quotes the formula as 

Hz = 1) + ^ (D " d) " Kd( ° ~ d ) cosec2Z 
where K ~ 0*57 

Z is the zenith angle of observation. 
Over large distances Z ^ 90° .'. cosec 2 Z ^ 1. 

Example 5.15 

If h x = 2 300 ft (at A), d = 46 miles 

h 2 = 1050 ft (at B), D = 84 miles 

h 3 = 1800 ft (at C). 

Can C be seen from /I ? 
By Eq. (5.32), 



„ 2 = 1800x46 + (g4 
84 

= 985-7 + 44-1 

= 1029-8 ft 



46) ( 



2300 
84 



0-57 x 46 



(5.32) 



(5.33) 



LEVELLING 



277 



2300' 




54-8' 



1800' 



Fig. 5.17 

The station C cannot be seen from A as h z is > 1029*8. 

If the line of sight is not to be nearer than 10 ft to the surface at 
B, then it would be necessary to erect a tower at C of such a height 
that the line of sight would be 10 ft above B, 

84 



i.e. so that its height h = (1 050 + 10 - 1 029*8) 

= 30*2 x 1*826 
~ 54* 8 ft 



46 



Exercises 5(e) (Intervisibility) 

19. Two ships A and B are 20 miles (32* 18 km) apart. If the observer 
at A is 20 ft (6* 096 m) above sea level, what should be the height of 
the mast of B above the sea for it to be seen at A? 

(Ans. 113* Oft (34* 5 m)) 

20. As part of a minor triangulation a station A was selected at 
708*63 ft A.O.D. Resection has been difficult in the area and as an 
additional check it is required to observe a triangulation station C 
35 miles away (reduced level 325*75 ft A.O.D.). If there is an inter- 
mediate hill at B, 15 miles from A (spot height shown on map near 

B 370 ft A.O.D.), will it be possible to observe station C from A as- 
suming that the ray should be 10 ft above B? 

(N.R.C.T. Ans. Instrument + target should be ~ 20 ft) 

21. (a) Discuss the effects of curvature and refraction on long sights 
as met with in triangulation, deriving a compounded equation for their 
correction. 

(b) A colliery headgear at A, ground level 452*48 ft A.O.D., is 
145ft to the observing platform. 

It is required to observe a triangulation station C, reduced level 
412*68 ft A.O.D., which is 15 miles from A, but it is thought that in- 
tervening ground at B approximately 500ft A.O.D. and 5 miles from A 



278 



SURVEYING PROBLEMS AND SOLUTIONS 



will prevent the line of sight. 

Assuming that the ray should not be nearer than 10 ft to the ground 
at any point, will the observation be possible? 

If not, what height should the target be at C? 

(R.I.C.S./M Ans. >5ft) 

22. Describe the effect of earth's curvature and refraction on long 
sights. Show how these effects can be cancelled by taking reciprocal 
observations. 

Two beacons A and B are 60 miles apart and are respectively 
120 ft and 1200 ft above mean sea level. At C, which is in the line 
AB and Is 15 miles from B,. the ground level is 548 ft above mean sea 
level. 

Find by how much, if at all, B should be raised so that the line 
of sight from A to B should pass 10 ft above the ground at C. The 
mean radius of the earth may be taken as 3960 miles. 

(L.U. Ans. +17 ft) 

5.65 Trigonometrical levelling 

For plane surveying purposes where the length of sight is limited 
to say 10 miles the foregoing principles can be applied, Fig. 5.18. 



— "I?* 




5280d 
tanoC 

1 

0-57of z 




1 


• 





Fig. 5.18 



The difference in elevation = h. 



h, = 5 280dtana+0'57d 

(5.34) 



where d = distance in miles 
a = angle of elevation. 
If the distance D is given in feet, then 

A, - A, - Dtana + 0-57 (^ 

= Dtana + 2'04 x 10~ 8 D 2 (5.35) 

N.B. It is considered advisable in trigonometrical levelling, and in 
normal geometrical levelling over long distances, to observe in both 
directions, simultaneously where possible, in order to eliminate the 
effects of curvature and refraction, as well as instrumental errors. 
This is known as reciprocal levelling. 



LEVELLING 279 

Example 5.16 The reduced level of the observation station A is 
350*36 ft A.O.D. From A, instrument height 4*31 ft, the angle of ele- 
vation is 5° 30' to station B, target height 6*44 ft. If the computed 
distance AB is 35 680* lft what is the reduced level of B? 

Reduced level of B = Reduced level of A + difference in eleva- 
tion + instrument height - target height 
By Eq. (5.35), 

Difference in elevation = 35 680*1 tan 5 ° 30' + 2*04 x 10~ 8 x 35 680' l 2 
- 3 435-60 + 25-97 
= 3 461- 57 ft 

Reduced level of B = 3 461*57 + 4*31 - 6- 44 

= 3 459- 44 ft A.O.D. 

Based on metric values the problem becomes: 

The reduced level of the observation station A is 106'790m A.O.D. 
From A, instrument height 1-314 m, the angle of elevation is 5° 30' to 
station B, target height 1*963 m. If the computed distance AB is 
10-875 3 km, what is the reduced level of B? 

Difference in elevation = 10 875*3 tan 5 ° 30' + 0*067 3 (10*875 3 2 ) 

= 1047-172 + 7-960 

= 1055- 132 m 

Reduced level of B = 1055-132 + 1-314 - 1*963 

= 1054* 483m (3 459* 59 ft) A.O.D. 

5.7 Reciprocal Levelling 

Corrections for curvature and refraction are only approximations 
as they depend on the observer's position, the shape of the geoid and 
atmospheric conditions. 

To eliminate the need for corrections a system of Reciprocal 
Levelling is adopted for long sights. 

In Fig. 5.19, 

Difference in level = d = fiX, = a^ + c + e-r-by 

from A = (a, - b x ) + (c - r) + e 

Also from B d = AX 2 = - (b 2 + c + e - r ~ a 2 ) 

= (« 2 ~b 2 ) - (c-r) - e 

By adding, 2d = (a, - ft,) + (a 2 - b 2 ) 

d = |[(a,-ft.) + (a 2 -b 2 )] 

(5.36) 



280 



SURVEYING PROBLEMS AND SOLUTIONS 



Uneo^g^f-^^^li^ 



Horizontal line 



Line due to refractTol? 

~-^-^e, 




H 



■fll 



b 2 



- L|n£_o'_sight_due to col/fmati, 



Horizontal line 



i : ;^Tw retraction 

?T^ Line due ^ ^^- 




Fig. 5.19 



Subtracting, 2(c - r + e) = [(a 2 - b 2 ) - (a, - b,)] 



Total error (c - r) + e = ±[(a 2 - b 2 ) - (a, - &,)] 



(5.37 



By calculating the error due to refraction and curvature, Eq. (5.28), 
for (c-r) the collimation error e may be derived. (See §5.4.) 

Example 5.17 (a) Obtain from first principles an expression giving 
the combined correction for earth's curvature and atmospheric refrac- 
tion in levelling, assuming that the earth is a sphere of 7 920 miles 
diameter. 

(b) Reciprocal levelling between two points Y and Z 
2 400ft apart on opposite sides of a river gave the following results: 





LEVELLING 






Instrument at 


Height of instrument 


Staff at 


Staff reading 


Y 


4-80 


Z 


5-54 


Z 


4-71 


y 


3-25 



281 



Determine the difference in level between Y and Z and the 
amount of any collimation error in the instrument. (I.C.E.) 

(a) By Eq. (5.28), c ~ 0'57d 2 

(b) By Eq. (5.36), 

Difference in level = |[(a, - ft,) + (a 2 - b 2 )] 

= i[(4-80 - 5-54) + (3-25 - 4*71)] 

= \ [-0-74 - 1-46] 

= - MO ft 
i.e. Z is MOft below Y 
By Eq. (5.37), 
Total error (c - r) + e = i [(a 2 ~ b z ) ~ (a, - &,)] 

= |[- 1*46 + 0-74] 









= 


- 0-36 ft 






By Eq. 


(5.28), (c- 
i.e 


-r) r^ 

e = 


0-57d 2 

. 57 /2400\ 2 
V5 280/ 

0-118 ft 

- 0-478 ft per 2 400 ft 

- 0-02 ft per 100 ft 












(collimation depressed) 


Cfcecfc 












Difference in 


i level = 


4-80 - 


- 5-54 - 0-48 + 0-12 = 


- MOft 


also 






3-25 


- 4-71 + 0-48 - 0-12 = 


- MOft 



5.71 The use of two instruments 

To improve the observations by removing the likelihood of clim- 
atic change two instruments should be used, as in the following 
example. 



282 



SURVEYING PROBLEMS AND SOLUTIONS 



Example 5.18 












Instrument 


Staff at 

A 


Mean 


Staff at 
B 


Mean 


Apparent Difference 
in level 


Remarks 


I L 


6-934 




9-424 






Inst. I on same 


M 


6-784 


6-784 


8-072 


8-073 


-1-289 


side as A 


U 


6-634 




6-722 








n L 


6-335 




6-426 






Inst. II on same 


M 


4-985 


4-984 


6-276 


6-276 


-1-292 


side as B 


U 


3-633 




6-126 








I L 


6-452 




6-514 






Inst. I on same 


M 


5-098 


5-099 


6-364 


6-364 


-1-265 


side as B 


U 


3-747 




6-214 








II L 


6-782 




9-249 






Inst. II on same 


M 6-632 


6-632 


7-893 


6-895 


-1-263 


side as A 


U 


1 6-482 




6-543 









4 ) -5-109 
True difference in level - 1*277 

Thus B is 1-277 ft below A. 



Exercises 5(f) (Reciprocal levelling) 

23. The results of reciprocal levelling between stations A and B 
1 500ft apart on opposite sides of a wide river were as follows: 



Level at 



Height of Eyepiece (ft) 



4-59 
4-37 



Staff Readings 



8-26 on B 
1-72 on A 



Find (a) the true difference in level between the stations 

(b) the error due to imperfect adjustment of the instrument assum- 
ing the mean radius of the earth 3 956 miles. 

(L.U./E Ans. (a) -3-16 ft; (b) + 0' 031 ft /100ft) 

24. In levelling across a wide river the following readings were taken: 



Instrument at 



Staff Reading at A 



5- 98 ft (l-823m) 
8- 20 ft (2- 499 m) 



Staff Reading at B 



8- 14 ft (2-481 m) 
10- 44 ft (3- 182 m) 



If the reduced level at A is 102*63 ft (31*282 m) above datum what 
is the reduced level of B? 

(Ans. 100*43 ft (30*612 m)) 



LEVELLING 



283 



5.8 Levelling for Construction 
5.81 Grading of constructions 

The gradient of the proposed construction will be expressed as 
1 in x, i.e. 1 vertical to x horizontal. 

The reduced formation level is then computed from the reduced 
level of a point on the formation, e.g. the starting point, and the pro- 
posed gradient. 

By comparing the existing reduced levels with the proposed re- 
duced levels the amount of cut and fill is obtained. 

If formation > existing, fill is required. 

If formation < existing, cut is required. 

Example 5.19 The following notes of a sectional levelling were taken 
along a line of a proposed road on the surface. 



B.S. 


I.S. 


F.S. 


Height of 
Collimation 


Reduced 
Level 


Horizontal 
Distance 


Remarks 


10-24 


4-63 
1-47 






104-52 



100 


B.M. 

Station 1 
Station 2 


8-52 


5-23 


0-41 






200 
300 


Station 3 
Station 4 


12-64 




3-37 
5-87 






400 
500 


Station 5 
Station 6 



Calculate the reduced level of each station and apply the conven- 
tional arithmetical checks. Thereafter calculate the depth of cutting 
and filling necessary at each station to form an even gradient rising at 
1 in 20 and starting at a level of 105 ft above datum at station 1. 

(M.Q.B./M) 



B.S. 


I.S. 


F.S. 


Height of 
Collimation 


Reduced 
Level 


Formation 
Level 


Cut 


Fill 


Station 


10-24 






114-76 


104-52 












4-63 






110-13 


105-00 


5-13 




1 




1-47 






113-29 


110-00 


3-29 




2 


8-52 




0-41 


122-87 


114-35 


115-00 




0-65 


3 




5-23 






117-64 


120-00 




2-36 


4 


12-64 




3-37 


132-14 


119-50 


125-00 




5-50 


5 






5-87 




126-27 
701-18 


130-00 




3-73 


6 


31-40 


11-33 


9-65 




9-65 


















21-75 



















284 



SURVEYING PROBLEMS AND SOLUTIONS 




NB. Vertical scale exaggerated 
(x10 horizontal scale) 



- 


CM 


CO 


* 


in 


Chainage 


0) 




* 
1 


8 




8 


Existing level 


o 
6 


o 


o 
6 

CM 


o 
to 

CM 


o 

8 


Formation level 


o> 

CM 
<*> 










Cut 




IS 

6 


<0 
CO 
CM 


o 
in 
in 


CO 
CO 


Fill 



Fig. 5.20 

Check 701-18 + 11*33 + 9*65 = 722-16 

(114*76 x 3) + (122-87 x 2) + (132-14 x 2) = 722-16 

5.82 The use of sight rails and boning (or travelling) rods 

Sight rails and boning rods are used for excavation purposes as- 
sociated with the grading of drains and sewers. 

The sight rails are established at fixed points along the excava- 
tion line, at a height above the formation level equal to the length of 
the boning rod. The formation level compared with the surface level 
gives the depth of excavation, Fig. 5.21. 



Boning rod 




Sight rail 



Fig. 5.21 



LEVELLING 



285 



When the boning rod is in line with the sight rails the excavation 
s at the correct depth, Fig. 5.22. 



Sight rail 



Boning rod 



^Jimmmmsmm^mmm^mm 



Length of 
boning rod 



Formation^ 



oradie_t_ 



Sight rail 



Length of 
boning rod 



Fig. 5.22 

Example 5.20 In preparing the fixing of sight rails, the following 
consecutive staff readings were taken from one setting of the level: 

Benchmark (165*65ft A.O.D.) 2*73 

Ground level at A 5*92 

Invert of sewer at A 10* 63 

Ground level at B 4*27 

Ground level at C 3*54 

If the sewer is to rise at 1 in 300 and the distance AB 105 ft and 
BC 153 ft, what will be the height of the sight rails for use with 10 ft 
boning rods? 

What is the reduced level of the invert at A, B and C? 



2-73 



165-65 
A.O.D. ! 



5-92 



(10-63 invert) 
1Q5' 



3-54 



153' 



168-38'A.O.D. 



Invert of •ew-rJJP. 



*mVmVMW;WM\WW/AWMWM«> 



300 



Fig. 5.23 
In Fig. 5.23, 

Height of collimation = 165*65 + 2* 73 
Invert of sewer at A = 168*38 - 10*63 
Sight rail at A = 157*75 + 10*00 

Ground level at A = 168*38 - 5*92 
Height of sight rail above ground at A 



168*38 ft A.O.D. 
157*75 ft 
167*75 ft 
162*46 ft 
5*29 ft 



286 



SURVEYING PROBLEMS AND SOLUTIONS 



Gradient of sewer 1 in 300 

Invert of sewer at B = Invert at A + rise due to gradient 

= 157-75 + 105/300 = 158- 10 ft 
Sight rail at B = 168* 10 ft 

Ground level at B = 168'38 - 4' 27 = 164- 11 ft 

Height of sight rail above ground at B = 3' 99 ft 

Invert of sewer at C = Invert at B + rise due to gradient 

= 158-10+153/300 = 158-61 ft 
Sight rail at C = 158-61 + 10-00 = 168*61 ft 

Ground level at C = 168-38 - 3' 54 = 164-84 ft 

3- 77 ft 

5. S3 The setting of slope stakes 

A slope stake is set in the ground at the intersection of the ground 
and the formation slope of the cutting or embankment. 

The position of the slope stake relative to the centre line of the 
formation may be obtained: 

(a) by scaling from the development plan, or 

(b) by calculation involving the cross-slope of the ground and the 
formation slope, using the rate of approach method suggested 
in §8.3. 



Formation level 




Fig. 5.24 

By the rate of approach method, in Fig. 5.24, 



w 



h < ■ A ° + w 



ft, = K 



2K 



(5.38) 
(5.39) 



LEVELLING 



287 



By Eq. (8.14), 


«*, - 


■*, 


1 1 






hi ~ k 


Similarly, 


d 2 - 


K 


l l 



(5.40) 
(5.41) 



M K 



Point A 



Point B 



5- 63 ft 



6- 13 ft 



Example 5.21 To determine the position of slope stakes, staff read- 
ings were taken at ground level as follows: 

Centre line of proposed road 
(Reduced level 103' 72 ft A.O.D.) 

50 ft from centre line and at 
right angles to it 

If the reduced level of the formation at the centre line is to be 
123-96 A.O.D., the formation width 20ft, and the batter is to be 1 in 2, 
what will be the staff reading, from the same instrument height at the 
slope stake and how far will the peg be from the centre line point A? 



123-96 




Fig. 5.25 

Gradient of AB = (6*13- 5*63) in 50ft i.e. 0-5 in 50ft 

1 in 100 

In Fig. 5.25, AA, = 123*96 - 103*72 

20 



By Eq. (5.38), XX, = fc = 20*24 + 



= 20*24 ft 
= 20*34 ft 



2x 100 
By Eq. (5.40), the horizontal distance <2,i.e. XP, 

h 20*34 



1 _ 1 

M K 



2 100 

100 x 20*34 
50 - 1 

= 41*51 ft 



288 SURVEYING PROBLEMS AND SOLUTIONS 

The distance from the centre line point A = 51* 51 ft 



the inclined length 


XP 


= 


41-51 x VUOO 2 + 1) 
100 






= 


41-72 ft 


Level of A 




= 


103-72 


Difference in level AP - 41 ' 5 } + 10 

100 




= 


0-52 ft 


Level of P 




= 


103-20 


Height of collimation = 103*72 + 5*63 




= 


109-35ft 


Staff reading at P 




= 


6- 15 ft 



Exercises 5(g) (Construction levelling) 

25. Sight rails are to be fixed at A and B 350 ft apart for the setting 
out of a sewer at an inclination of 1 in 200 rising towards B. 

If the levels of the surface are A 106*23 and B 104*46 and the 
invert level at A is 100*74, at what height above ground should the 
sight rails be set for use with boning rods 10 ft long? 

(Ans. 4*51 at A; 8'03 at B) 

26. A sewer is to be laid at a uniform gradient of 1 in 200 between 
two points X and Y, 800 ft apart. The reduced level of the invert at 
the outfall X is 494*82. 

In order to fix sight rails at X and Y, readings are taken with a 
level in the following order: 

Reading Staff Station 

B.S. 2-66 T.B.M. (near X) Reduced level 504-64 

I.S. a Top of sight rail at X 

I.S. 3-52 Peg at X 

F.S. 1-80 T.P. between X and Y 

B.S. 7-04 T.P. between X and Y 

I.S. b Top of sight rail at Y 

F.S. 6-15 Peg at Y 

(i) Draw up a level book and find the reduced levels of the pegs, 
(ii) If a boning rod of length 9'- 6" is to be used, find the readings 

a and b. 
(iii) Find the height of the sight rails above the pegs at X and Y. 
(L.U. Ans. (ii)2*98, 4*22; (iii) 0- 54, 1*93) 

27. The levelling shown on the field sheet given below was under- 
taken during the laying out of a sewer line. Determine the height of 



LEVELLING 



289 



the ground at each observed point along the sewer line and calculate 
the depth of the trench at points X and Y if the sewer is to have a 
gradient of 1 in 200 downwards from A to B and is to be 4*20 ft below 
the surface at A. 



B.S. 


I.S. 


F.S. 


Distance (ft) 


Remarks 


11-21 








B.M. 321-53 


4-56 


3-78 


5-82 



100 




11-65 




3-66 


200 


Point X 


2-40 




3-57 


300 




7-82 


5-91 


10-81 


400 
500 






6-56 




600 


Point Y 


6-32 




8-65 
3-81 


700 


Point B 
B.M. 329-15 



(R.I.C.S./M/L Ans. 6*10, 9'03) 



Exercises 5(h) (General) 



28. The following staff readings in fact were taken successively with 
a level, the instrument having been moved forward after the second, 
fourth and eighth reading. 1*54, 7*24, 4'03, M5, 8'62, 8'52, 6*41, 
1-13, 7-31, 2-75 and 5'41. 

The last reading was taken with the staff on a bench mark having 
an elevation of 103-74 ft. 

Enter the readings in level book form, complete the reduced levels 
and apply the usual checks. 

29. The following readings were taken using a dumpy level on a 
slightly undulating underground roadway. 



B.S. 


I.S. 


F.S. 


Reduced Level 


Distance 


Remarks 


5-32 


6-43 
5-23 




+ 8 752-20 



100 
200 


Point A + 8 752-20 


3-06 


4-30 
2-23 


4-12 




300 
400 
500 




3-02 


4-01 
5-12 


1-09 




600 
700 
800 








6-67 




900 


Point B 



Work out the reduced levels relative to the assumed datum of mean 



290 



SURVEYING PROBLEMS AND SOLUTIONS 



sea level +10 000 ft (as used by the National Coal Board to avoid 
negative reduced levels). 

State the amount of excavation necessary at point B to form an 
even gradient dipping 1 in 300 from A to B, the reduced level of A 
to remain at 8 752*20 ft. 

(Ans. 2- 52 ft) 

30. Reduce the page of a level book and plot the result to a scale of 
1 = 100' horizontal and 1" = 10' vertical. 



B.S. 


I.S. 


F.S. 


Rise 


Fall 


Reduced Level 


Distance 


Remarks 


9-92 


8-22 
5-98 








25-23 






100 


B.M. 

Start of Section 


7-35 


5-59 


3-15 








200 
300 




8-13 


6-05 
5-63 
5-00 


2-12 








400 
460 
500 
560 




5-63 


4-19 
5-91 


3-65 








600 
630 
700 




4-71 


5-35 
4-01 


8-04 








800 
830 
900 




6-24 


5-82 
4-36 


2-73 
3-72 








1000 
1100 
1200 


End of Section 
B.M. 

(R.I.C.S./Q) 



31. A levelling party ran a line of levels from point A at elevation 
135*43 to point B for which the reduced level was found to be 87*15. 
A series of flying levels (as below) was taken back to the starting 
point A. 



B.S. 


F.S. 


Remarks 


9-67 




B 


11-54 


1-38 




8-22 


4-81 




7-94 


3-35 




10-56 


2-07 




9-92 


5-33 




8-88 


1-04 






0-42 


A 



LEVELLING 



291 



Find the misclosure on the starting point. 

(L.U./E Ans. 0'05ft) 

32. (a) Explain the difference between 'rise and fall' and 'height of 
collimation' method of reducing levels, stating the advantages and dis- 
advantages of each. 

(b) The following is an extract from a level book. Reduce the 
levels by whichever method you think appropriate, making all the 
necessary checks and insert the staff readings in the correct blank 
spaces for setting in the levels pegs A, B and C so that they have 
the reduced levels given in the book. 



B.S. 


I.S. 


F.S. 


Reduced Level 


Distance 


Remarks 


3-24 






58-63 




B.M. 1 


5-03 




9-65 






C.P. 




6-42 









Beginning of Se 




9-69 






50 






- 




49-69 


50 


Peg A 


4-19 




10-87 




100 


C.P. 




- 




48-55 


100 


Peg B 




5-54 






150 






- 




47-41 


150 


Peg C 




4-30 






200 


End of Section 


11-73 




2-32 






C.P. 






8-61 


51-35 




B.M. 2 



(R.I.C.S./M/L Ans. Staff Readings, A 7'56, B 2*02, C 3-16; 

Error in levelling, 0*02) 

33. The level book refers to a grid of levels taken at 100 ft intervals 
on 4 parallel lines 100ft apart. 

(A) Reduce and check the level book. 

(B) Draw a grid to a scale of 50 ft to lin. and plot the contours 
for a 2 ft vertical interval. 



B.S. 


I.S. 


F.S. 


Reduced Level 


Distance on Line 


Remarks 


1-23 


2-75 
3-51 
4-26 




89-14 



100 
200 
300 


Line A T.B.M. 




12-35 






300 


Line B 




9-06 






200 






6-78 






100 






4-18 











4-15 


5-51 


6-97 





100 


Line C 



292 



SURVEYING PROBLEMS AND SOLUTIONS 



B.S. 


LS. 


F.S. 


Reduced Level 


Distance on Line 


Remarks 




7-88 






200 






10-45 






300 






8-93 






300 


Line D 




7-18 






200 






5-34 






100 






4-59 











8-37 




2-62 
4-14 






C.P. 
T.B.M. 



(R.I.C.S./L/M) 

34. Discuss the various ways in which 'errors' can occur in levelling 
and measures that can be adopted to keep each source of error to a 
minimum. 

In levelling from a bench mark 347*79 ft above O.D. and closing on 
to another 330*61 ft above O.D., staff readings were taken in the fol- 
lowing order: 

3*72, 8*21; 4*91, 8*33, 7*28; 0*89, 4*27; 2*28, 3*91, 3*72, 9*23. 

The position of the instrument was moved immediately after taking 
the 2nd, 5th, and 7th readings indicated by semi-colons in the above 
series of readings. 

Show how these readings would be booked and the levels reduced 
using either the 'collimation' or the 'rise and fall' method. Carry out 
the usual arithmetical checks and quote the closing error. 

Explain briefly why it is particularly important not to make a mis- 
take in reading an intermediate sight. /T _ „ N 

(l.U.L.) 

35. The record of a levelling made some years ago has become of cur- 
rent importance. Some of the data are undecipherable but sufficient re- 
main to enable all the missing values to be calculated. Reproduce the 
following levelling notes and calculate and insert the missing values. 



B.S. 


I.S. 


F.S. 


Rise 


Fall 


Reduced Level 


Remarks 


2-36 








1-94 


121-36 


B.M. at No. 1 Shaft 


4-05 


4-31 
6-93 

3-22 
6-53 


7-29 
7-79 


1-58 


4-46 
0-63 


113-32 
112-01 


B.M. on School 



LEVELLING 



293 



B.S. 


I.S. 


F.S. 


Rise 


Fall 


Reduced Level 


Remarks 






5-86 




3-10 
21-18 


113-53 


B.M. on Church 


14-96 





36. The following is an extract from a level book 



B.S. 


I.S. 


F.S. 


Reduced Level 


Remarks 


4-20 


2-70 






Point A 


2-64 


3-42 

9-51 

11-74 


11-40 


119-30 


C.P. B.M. 119-30 


2-56 


3-10 
6-91 


13-75 




C.P. B 


3-61 


5-60 


11-23 




C.P. C 


12-98 




3-61 




C.P. C 


13-62 




3-31 




C.P. B 


12-03 




2-51 
4-83 




C.P. B.M. 
Point A 



(a) Reduce the above levels. 

(b) If you consider a mistake has been made suggest how it 
occurred. 

(c) Give reasons for choice of 'Rise and Fall' or 'Height of Col- 
limation' for reducing the levels. The B.S. and F.S. lengths 
were approximately equal. 

(L.U. Ans. probably 11*98 instead of 12*98) 

37. The following are the levels along a line ABC. 



Distance 


Reduced Level 


Remarks 





0-00 


At A 


10 


1-21 




20 


2-46 




30 


3-39 




40 


4-54 


At B 


50 


6-03 




60 


7-65 




70 


9-03 




80 


10-32 


At C 



294 



SURVEYING PROBLEMS AND SOLUTIONS 



Plot the reduced levels to a scale of 10 ft to lin. for the horizon- 
tal scale and 1ft to lin. for the vertical scale. 

A roadway is to be constructed from A to C at a uniform gradient. 
From the section state the height of filling required at each plotting 
point. (R.I.C.S./M) 

38. In order to check the underground levellings of a colliery it was 
decided to remeasure the depth of the shaft and connect the levelling 
to a recently established Ordnance Survey Bench Mark A, 272*45 ft 
above O.D. 

The following levels were taken with a dumpy level starting at A 
to the mouth of the shaft at D. 



B.S. 


F.S. 


Reduced Level 


Remarks 


2-17 




272-45 ft 


B.M. at A 


3-36 


11-32 




Mark B 


5-79 


7-93 




Mark C 




0-00 




Mark D on rails 



The vertical depth of the shaft was then measured from D to E 
at the pit bottom and found to be 1 745 ft 8^ in. 

A backsight underground to E was found to be 3*98 and a fore- 
sight to the colliery Bench Mark F on a wall near the pit bottom was 
2-73. 

Tabulate the above readings and find the value of the underground 
B.M. at F expressing this as a depth below Ordnance Datum. 

(Ans. 1479-94 ft) 
39. Levels were taken at 100 ft intervals down a road with a fairly 
uniform gradient and the following staff readings booked: 



B.S. 


I.S. 


F.S. 


Distance 




7-00 


9-50 





100 


T.B.M. 98-50 


11-75 


8-55 
10-81 


6-00 


200 
300 
400 




13-05 


7-90 
10-20 


5-60 


500 
600 
700 




12-70 


8-48 
10-98 
13-20 


6-00 


800 

900 

1000 

1 100 





LEVELLING 295 

Errors were made in booking, correct these and reduce the levels. 

(L.U. Ans. B.S. and F.S. transposed) 

40. Spot levels are given below at 200 ft intervals on a grid ABCD. 
Draw the plan to a scale of 200 ft to 1 in. and show on it where you 
would place the level in order to take readings. 

Draw up a level book by the 'height of collimation' method show- 
ing your readings. Take the level at A as a T.B.M. 

A 100-70 102-00 103-50 105*20 106-80 108-20 109-50 B 

101-30 103-40 104-10 106'30 108*20 109-30 110-70 

105-00 106-20 107-30 109-10 110-40 111-50 112*30 

D 108-00 107-10 108-60 110-40 111-30 112*20 113-80 C 

(L.U.) 

41. In levelling up a hillside, the sight lengths were observed with 
stadia lines, the average length of the ten backsights and foresights 
being 70 ft and 35 ft respectively. 

Since the observed difference of the reduced level of 78* 40 ft was 
disputed, the level was set up midway between two pegs A and B 
300 ft apart, and the reading on A was 4*60 and on B 5*11; and when 
set up in line AB, 30 ft behind B, the reading on A was 5*17 and on 
B 5-64. 

Calculate the true difference of reduced level. 

(L.U. Ans. 78-35 ft; 0*013 ft per 100ft) 

42. A, B, C, D, E and F are the sites of manholes, 300 ft apart on a 
straight sewer. The natural ground can be considered as a plane sur- 
face rising uniformly from A to F at a gradient of 1 vertically in 500 
horizontally, the ground level at A being 103*00. The level of the 
sewer invert is to be 95*00 at A, the invert then rising uniformly at 

1 in 200 to F. Site rails are to be set up at A, B, C, D, E and F so 
that a 10ft boning rod or traveller can be used. The backsights and 
foresights were made approximately equal and a peg at ground level at 
A was used as datum. 

Draw up a level book showing the readings. (L.U.) 

43. The following staff readings were obtained when running a line of 
levels between two bench marks A and B: 

3*56 (A) 6*68, 7-32, 9*89 change point, 2*01, 6*57, 7*66, C.P. 
5*32, 4*21, 1-78, C.P. 4*68, 5*89, 2*99 (S) 

Enter and reduce the readings in an accepted form of field book. 
The reduced levels of the bench marks at A and B were known to be 
143-21 ft and 136*72 ft respectively. 

It is found after the readings have been taken with the staff 



296 



SURVEYING PROBLEMS AND SOLUTIONS 



supposedly vertical, as indicated by a level on the staff, that the level 
is 5 ° in error in the plane of the staff and instrument. 

Is the collimation error of the instrument elevated or depressed 
and what is its value in seconds if the backsights and foresights 
averaged 100ft and 200 ft respectively. 

(L.U. True difference in level 6*72; collimation elevated 119sec) 

44. Undernoted are levels taken on the floor of an undulating under- 
ground roadway AB, 10 ft in width and 6 ft in height, which is to be 
regraded and heightened. 



Distance 


(ft) 


B.S. 


I.S. 


F.S. 


Rise 


Fall 


Reduced Level 


Remarks 







6-95 










30-00 


Floor level at A 


50 






3-40 












100 






0-65 












150 




2-50 




0-00 










200 






3-50 












250 






6-00 












300 






6-50 












350 




7-75 




5-00 










400 






4-50 












450 






1-25 












500 




1-00 


2-20 


6-65 
6-55 








Floor level at B 
Floor level at A 



Plot the section of the roadway to a scale of lin. = 50 ft for hori- 
zontals and lin. = 10ft for verticals. Thereafter calculate and mark 
on the section the amount of ripping and filling at the respective 50 ft 
intervals to give a uniform gradient from A to B and a minimum height 
of 8 ft. 

Calculate the volume, in cubic yards, of the material to be ripped 
from the roof in giving effect to the above improvements. 

(M.Q.B./M Ans. 320yd 3 ) 

45. The centre-line of a section of a proposed road in cutting is indi- 
cated by pegs at equal intervals and the corresponding longitudinal 
section gives the existing ground level and the proposed formation 
level at each peg, but no cross-sections have been taken, or sidelong 
slopes observed. 

Given the proposed formation width (d) and the batter of the sides 
(S horizontal to 1 vertical) how would you set out the batter pegs 
marking the tops of the slopes at each centre line peg, without taking 
and plotting the usual cross-sections? 

An alternative method would be acceptable. (I.C.E.) 



LEVELLING 



297 



46. (a) Determine from first principles the approximate distance at 
which correction for curvature and refraction in levelling amounts to 
0*01 ft, assuming that the effect of refraction is one seventh that of 
the earth's curvature and that the earth is a sphere of 7 920 miles di- 
ameter. 

(b) Two survey stations A and B on opposite sides of a river 
are 2510 ft apart, and reciprocal levels have been taken between them 
with the following results: 



Instrument at 


Height of instrument 


Staff at 


Staff reading (ft) 


A 
B 


4-83 
4-91 


B 

A 


6-02 
3-98 



Compute the ratio of refraction correction to curvature correction, 
and the difference in level between A and B. 

(LOE. Ans. (a) ~ 700ft; 

(b) A is 1-06 ft above B. 
Ratio ~ 0*14 to 1) 



Bibliography 

HOLLAND, J.L., WARDELL, K. and WEBSTER, A.G., Surveying, Coal 

Mining Series (Virtue). 

SANDOVER, J. A. Plane Surveying (Edward Arnold). 

TRUTMAN, O., Levelling (Wild (Heerbrugg)). 

PARRY, R. and JENKINS, W.R., Land Surveying (Estates Gazette). 

BANNISTER, A. and RAYMOND, S., Surveying (Pitman). 

CURTEN, w. and LANE, R.F., Concise Practical Surveying (English 

Universities Press). 



O TRAVERSE SURVEYS 

The purpose of traverse surveys is to control subsequent detail, 
i.e. the fixing of specific points to which detail can be related. The 
accuracy of the control survey must be superior to that of the subsi- 
diary survey. 

A traverse consists of a series of related points or stations, which 
when connected by angular and linear values form a framework. 

6.1 Types of Traverse 
6.11 Open 

Traverse ABCDE, Fig.6.1. The start and finish are not fixed 
points. 




c 
Fig. 6. 1 Open traverse 

A check on the angles may be made by (a) taking meridian observa- 
tions at the start and finish or (b) taking observations to a common 
fixed point X. 

6.12 Closed 

(a) On to fixed points. If the start and finish are fixed points, e.g. 
A and E, then the length and bearing between them is known. From 
the traverse the distance may be computed. 

(b) Closed polygon, Fig. 6.2. 

Checks (i) The sum of the deflection angles should equal 360°, i.e. 
2a = a, + a 2 + a 3 ...a n = 360° = 4 x 90° 

or (ii) The sum of the internal angles should equal 

(2« - 4) x 90° 
where n = no. of angles or sides, i.e. 

S/3 = jQ, + /S 2 + p 3 .../3 n = (2n -4)90° (6.1) 

or (iii) The sum of the external angles should equal 

298 



TRAVERSE SURVEYS 299 

(2n + 4) 90° 
Le - 20 = 0, + 2 + 3 ...0 n = (2n + 4)90° (6.2) 




«4 

Fig. 6.2 Closed traverse 



N.B. a, + /3, = a 2 + 2 = a n + n = 2 x 90° 

2(a + 0) = (4 x 90) + (2n - 4)90 = (2 x 90) x n. 

The sum will seldom add up exactly to the theoretical value and 
the 'closing error' must be distributed before plotting or computing. 

6.2 Methods of Traversing 

The method is dependent upon the accuracy required and the equip- 
ment available. The following are alternative methods. 

(1) Compass traversing using one of the following: 

(a) a prismatic compass 

(b) a miners' dial 

(c) a tubular or trough compass fitted to a theodolite 

(d) a special compass theodolite. 

(2) Continuous azimuth (fixed needle traversing) using either (a) 
a miners' dial or (b) a theodolite. 

(3) Direction method using any angular measuring equipment. 

(4) Separate angular measurement (double foresight method) using 
any angular measuring equipment. 



300 



SURVEYING PROBLEMS AND SOLUTIONS 



6.21 Compass traversing (loose needle traversing), Fig. 6.3 

Application. Reconnaissance or exploratory surveys. 

Advantages. (1) Rapid surveys. 

(2) Each line is independent — errors tend to compensate. 

(3) The bearing of a line can be observed at any point 
along the line. 

(4) Only every second station needs to be occupied (this 
is not recommended because of the possibility of 
local attraction) 

Disadvantages. (1) Lack of accuracy. (2) Local attraction. 
Accuracy of survey. Due to magnetic variations, instrument and obser- 
vation errors, the maximum accuracy is probably limited to ± lOmin, 
i.e. linear equivalent 1 in 300. 

Detection of effects of local attraction. Forward and back bearings 
should differ by 180° assuming no instrumental or personal errors exist. 
Elimination of the effect of local attraction. The effect of local attrac- 
tion is that all bearings from a given station will be in error by a con- 
stant value, the angle between adjacent bearings being correct. 

Where forward and back bearings of a line agree this indicates that 
the terminal stations are both free of local attraction. 

Thus, starting from bearings which are unaffected, a comparison of 
forward and back bearings will show the correction factors to be applied. 




X source of 
attraction 



Fig. 6.3 Compass traversing 

In Fig. 6.3 the bearings at A and B are correct. The back bearing 
of CB will be in error by a compared with the forward bearing BC. 



TRAVERSE SURVEYS 



301 



The forward bearing CD can thus be corrected by a. Comparison of 
the corrected forward bearing CD with the observed back bearing DC 
will show the error /3 by which the forward bearing DA must be cor- 
rected. This should finally check with back bearing AD. 



Example 6.1 










Line 


Forward Bearing 


Back Beari 




AB 


120° 10' 


300° 10' 




BC 


124° 08' 


306° 15' 




CD 


137° 10' 


310° 08 ' 




DE 


159° 08' 


349° 08' 




EF 


138° 15' 


313° 10' 



Station 


Back Bearing 


Forward Bearing 


Correction 


Corrected 
Forward 
Bearing 


± 180° 


A 


300° 10 ' 


120° 10* 
124° 08' 


- 




B 


124° 08' 


304° 08' 


C 


306° 15' 


137° 10' 


-2° 07' 


135° 03' 


315° 03' 


D 


310° 08' 


159° 08' 


+ 4° 55' 


164° 03' 


344° 03 ' 


E 


349° 08' 
313° 10' 


138° 15' 


-5° 05' 


133° 10 ' 


313° 10' 


F 





Thus stations A, B, and F are free from local attraction. 

N.B. (1) Line AB. Forward and Back bearings agree; therefore sta- 
tions A and B are free from attraction 

(2) Corrected forward bearing at B + 180° compared with back 
bearing at C shows an error of 2° 07', i.e. 304° 08' - 
306° 15' = -2° 07'. 

(3) Corrected forward bearing CD 137° 10' - 2° 07' = 135° 03'. 

(4) Comparison of corrected forward bearing EF + 180° agrees 
with back bearing FE. Therefore station F is also free 
from local attraction 



6.22 Continuous azimuth method (Fig. 6.4) 

This method was ideally suited to the old type of miners' dial with 
open-vane sights which could be used in either direction. 

The instrument is orientated at each station by observing the back- 
sight, with the reader clamped, from the reverse end of the 'dial' sights. 

The recorded value of each foresight is thus the bearing of each 
line relative to the original orientation. For mining purposes this was 
the magnetic meridian and hence the method was known as 'the fixed 
needle method*. 



302 



SURVEYING PROBLEMS AND SOLUTIONS 



The method may be modified for use with a theodolite by changing 
face between backsight and foresight observations. 




Fig. 6.4 Continuous azimuth method of traversing 

6.23 Direction method 

The continuous azimuth method of traversing is restrictive for use 
with the theodolite where the accuracy must be improved, as the obser- 
vations cannot be repeated. 

To overcome this difficulty, and still retain the benefit of carrying 
the bearing, the Direction method may be employed. 

This requires only approximate orientation, corrections being made 
either on a 'Direction' bearing sheet if one arc on each face is taken, 
or, alternatively, on the field booking sheet. 

N.B. No angles are extracted, the theodolite showing approximate 
bearings of the traverse lines as the work proceeds. 
Direction bearing sheet 



Set! 


Obs. 


at 


to 


A 


E 




B 


B 


A 




C 


C 


B 




D 


D 


C 




E 


E 


D 




A 



Mean observed 
directions 
08-00 
283 09.05 
103 00.60 
345 37*05 
165 36-40 
039 40-05 
219 55-50 
101 31-35 
281 31-20 
180 15-10 



Corr- 
ection 



+ 8-45 
+ 8-45 
+ 9-10 
+ 9-10 
-6-35 
-6-35 
-6-20 
-6-20 



Back 
bearing 
08-00 

103 09-05 

165 45-50 

219 49-15 

281 25-00 



Initial bearing 
j Error 



Forward 
bearing 

283 09-05 

345 45-50 

039 49-15 

101 25-00 

180 08-90 

180 08-00 

0-90 



Final 
correction 

-0-18 

-0-36 

-0-54 

-0-72 

-0-90 



Final 
bearing 
08-00 
283 08-87 



345 45-14 



039 48-61 



101 24-28 



180 08-00 



TRAVERSE SURVEYS 



303 



N.B. (1) Here the instrument has been approximately orientated at 
each station, i.e. the reciprocal of the previous forward bearing is set 
as a back bearing. Any variation from the previous mean forward bear- 
ing thus requires an orientation correction. 

(2) In the closed traverse the closing error is seen immediately 
by comparing the first back bearing with the final forward bearing. 

(3) As a simple adjustment the closing error is distributed 
equally amongst the lines. 

Method of booking by the direction method 







Station set at A 




Back bearing AE C 


°08' 


Arc 


Obs. 
to 


FL. 


F.R. 


Mean 


Correction 


Back 
bearing 


Forward 
bearing 






o ' 


o ' 


' 


' 


' 


' 


1 


E 


08-0 


180 08-0 






08-00 






B 


283 09-0 


103 09-1 








283 09-05 


2 


E 


090 12-2 


270 12-2 


090 12-20 


-90 04-20 


08-00 






B 


013 13-2 


193 13'4 


013 13-30 


-90 04-20 




283 09-10 


3 


E 


27-6 


27-4 


27-50 


19-50 


08-00 






B 


28-5 


28*5 


28-50 


- 19-50 




283 09-00 



Station set at B 



Mean Forward Bearing 283 09-05 



Back bearing BA 103° 09-05* 



Arc 


Obs. 
to 


F.L. 


F.R. 


Mean 


Correction 


Back 
bearing 


Forward 
bearing 






' 


' 


' 


' 


' 


' 


1 


A 


103 00-6 


283 00-5 


103 00-55 


+ 08-50 


103 09-05 






C 


345 37-1 


165 37*2 


345 37-15 


+ 08-50 




345 45-65 


2 


A 


199 04-9 


019 05-0 


199 04-95 


-95 55-90 


103 09-05 






C 


081 41-3 


261 41-5 


081 41-40 


-95 55-90 




345 45-50 


3 


A 


15-6 


15-7 


15-65 


- 06-60 


103 09-05 






C 


52-0 


52-2 


52-10 


06-60 




345 45-50 



Mean Forward Bearing 345 45-55 



N.B. 



(1) No angles are extracted. 

(2) The back bearing is approximately set on the instrument for 
the backsight, e.g. AE is set exactly here 0°08-0' 

.*• AB is the correctly oriented bearing 283° 09-0' . 

(3) On arc 2 the instrument zero is changed. After taking out 
the mean of the faces, a correction is applied to give the back bearing, 
i.e. 090° 12-20' 

- 0° 08-00' 
Correction 090° 04-20' 
This correction is now applied to give the second forward bearing. 



304 SURVEYING PROBLEMS AND SOLUTIONS 

(4) On the 3rd and subsequent arcs, if required, only the min- 
utes are booked, a new zero being obtained each time. 

(5) The mean forward bearing is now extracted and carried for- 
ward to the next station. 

6.24 Separate angular measurement 

Angles are measured by finding the difference between adjacent 
recorded pointings. 

After extracting the mean values these are converted into bearings 
for co-ordinate purposes. 

In the case of a closed polygon, the angles may be summated to 
conform with the geometrical properties, Eq. (6.1) or (6.2). 

Exercises 6(a) 

1. A colliery plan has been laid down on the national grid of the 
Ordnance Survey and the co-ordinates of the two stations A and B 
have been converted into feet and reduced to A as a local origin. 

Departure (ft) Latitude (ft) 
Station 4 

Station B East 109-2 South 991-7. 
Calculate the Grid bearing of the line AB. The mean magnetic 
bearing of the line AB is S 3° 54 'W and the mean magnetic bearing of 
an underground line CD is N 17°55'W. 
State the Grid bearing of the line CD. 

(M.Q.B./M Ans. 331° 54') 

2. The following angles were measured in a clockwise direction, from 
the National Grid North lines on a colliery plan: 

(a) 156° 15' (b) 181° 30' (c) 354° 00' (d) 17° 45' 
At the present time in this locality, the magnetic north is found to 
be 10° 30 'W of the Grid North. 

Express the above directions as quadrant bearings to be set off 
using the magnetic needle. 

(M.Q.B./UM Ans. (a)S13°15'E; (b)S12°00'W; 
(c)N4°30'E; (d)N28°15'E) 

3. The following underground survey was made with a miners' dial in 
the presence of iron rails. Assuming that station A was free from local 
attraction calculate the correct magnetic bearing of each line. 

Station BS FS 

A 352° 00' 

B 358° 30' 12° 20' 

C 14° 35' 282° 15' 



TRAVERSE SURVEYS 305 

Station BS FS 

D 280° 00' 164° 24' 
E 168° 42' 200° 22' 
(Ans. 352°00'; 05°50'; 273°30'; 157°54'; 189°34') 
(N.B. A miners' dial has vane sights, i.e. B.S. = F.S., not reciprocal 
bearings). 

4. The geographical azimuth of a church spire is observed from a tri- 
angulation station as 346° 20'. At a certain time of the day a magnetic 
bearing was taken of this same line as 003° 23'. On the following day 
at the same time an underground survey line was magnetically observed 
as 195° 20' with the same instrument. 

Calculate (a) the magnetic declination, 

(b) the true bearing of the underground line. 

(Ans. 17°03'W; 178° 17') 

5. Describe and sketch a prismatic compass. What precautions are 
taken when using the compass for field observations? 

The following readings were obtained in a short traverse ABC A. 
Adjust the readings and calculate the co-ordinates of B and C if the 
co-ordinates of A are 250 ft E, 75 ft N. 

Line Compass bearing Length (ft) 

AC 00° 00' 195-5 

AB 44° 59' 169-5 

BA 225° 01' 169-5 

BC 302° 10' 141-7 

CB 122° 10' 141-7 

CA 180° 00' 195-5 

(R.I.C.S. Ans. B 370E, 195N C 250E, 270N) 

6. The following notes were obtained during a compass survey made 
to determine the approximate area covered by an old dirt-tip. 

Correct the compass readings for local attraction. Plot the survey 
to a scale of 1 in 2400 and adjust graphically by Bowditch's rule. 
Thereafter find the area enclosed by equalising to a triangle. 

Line Forward bearing Back bearing Length (ft) 



AB 


N57°10'E 


S 58°20'W 


750 


BC 


N81°40'E 


S 78°00'W 


828 


CD 


S 15°30'E 


N 15°30'W 


764 


DE 


S 10°20'W 


N 12°00'E 


405 


EF 


S 78°50'W 


N76°00'E 


540 


FG 


N69°30'E 


S 68°30'W 


950 


GA 


N 22° 10'W 


S 19°30'E 


383 



(N.R.C.T. Ans. AB 54°40'; BC 78°00'; CD 164° 30'; D£190°20'; 
EF257°10'; FG 291° 40'; GA 338° 00'; area 32-64 acres) 



306 



SURVEYING PROBLEMS AND SOLUTIONS 



7. A and B are two reference stations in an underground roadway, 
and it is required to extend the survey through a drift from station B 
to a third station C. The observations at B were as follows: 

Horizontal angle ABC 271° 05' 20". 

Vertical angle to staff at C + 10° 15' 00". 

Staff reading at C 1-50 ft. 

Instrument height at B 5ft 7 in. 

Measured distance BC 284-86 ft. 
The bearing of AB was 349° 56' 10" and the co-ordinates of B 
E 4689-22 ft, N 5873-50 ft. 

Calculate the true slope of BC to the nearest 10 seconds, the 
horizontal length of BC, its bearing, and the co-ordinates of C. 
(N.R.C.T. Ans. 11°06'20"; 279-55 ft; 
081°01'30"; E 4965-35, N 5917-11) 



6.31 



6.3 Office Tests for Locating Mistakes in Traversing 
A mistake in the linear value of one line 



If the figure is closed the co-ordinates can be computed and the 
closing error found. 




Linear mistake 



Fig. 6.5 Location of a linear mistake 

Let the computed co-ordinates give values for ABC X D^A V Fig. 6.5. 

The length and bearing of AA y suggests that the mistake lies in 
this direction, and if it is parallel to any given line of the traverse 
this is where the mistake has been made. 

The amount AA X is therefore the linear mistake, and a correction 
to the line BC gives the new station values of C, D and thus closes 
on A. 

If the closing error is parallel to a number of lines then a repetition 
of their measurements is suggested. 



TRAVERSE SURVEYS 

6.32 A mistake in the angular value at one station 

Let the traverse be plotted as ABCD^A V Fig. 6.6 



307 




Station where 
mistake occurred 



Fig. 6.6 Location of an angular mistake 



The closing error AA y is not parallel to any line but the perpendi- 
cular bisector of /L4, when produced passes through station C. Here 
an angular mistake exists. 

Proof. AA) represents a chord of a circle of radius AC, the perpendi- 
cular bisector of the chord passing through the centre of the circle of 
centre C. 

The line CD, must be turned through the angle a = ACA V 

6.33 When the traverse is closed on to fixed points and a mistake in 
the bearing is known to exist 

The survey should be plotted or computed from each end in turn, 
i.e. ABCDE - E,D X C,BA, Fig. 6.7. 

The station which is common to both systems will suggest where 
the mistake has been made. 




Fig. 6.7 



308 SURVEYING PROBLEMS AND SOLUTIONS 

If there are two or more mistakes, either in length or bearing, then 
it is impossible to locate their positions but they may be localised. 

6.4 Omitted Measurements in Closed Traverses 

Where it is impossible to measure all the values (either linear, 
angular or a combination of both) in a closed traverse, the missing 
quantities can be calculated provided they do not exceed two. 

As the traverse is closed the algebraic sum of the partial co-ordin- 
ates must each sum to zero, i.e. 

/,sin0, + / 2 sin0 2 + / 3 sin0 3 + ... / n sin0 n = 
/,cos0, + f 2 cos0 2 + J 3 cos0 3 + ... I n cos0 n = 

where the lengths of the lines are /,/ 2 / 3 etc -» ana< the bearings 0,0 2 3 
etc. 

As only 2 independent equations are involved only 2 unknowns are 
possible. 

Failure to close the traverse in any way transfers all the traverse 
errors to the unknown quantities. Therefore use of the process is to be 
deprecated unless there is no other solution. 

Six cases may arise: 

(1) Bearing of one line. 

(2) Length of one line. 

(3) Length and bearing of one line. 

(4) Bearing of two lines. 

(5) Length of two lines. 

(6) Bearing of one line and length of another line. 

Cases 1, 2 and 3 are merely part of the calculation of a join between 
two co-ordinates. 

6.41 Where the bearing of one line is missing 

/ n sin0 n = P (1) where P = the sum of the other partial 

departures 

/ n cos d n = Q (2) where Q = the sum of the other partial 

latitudes 

Dividing (1) by (2), 

P AE 

tan0 n = — i.e. jrzi = the difference in the total co- 

^ ordinates of the stations form- 

ing the ends of the missing 
line (6.3) 



TRAVERSE SURVEYS 309 

6.42 Where the length of one line is missing 

/ n sin0 n = P (1) 

/ n cos0 n = Q (2) 

By squaring each and adding 

/ 2 sin 2 n = P 2 
/ n 2 cos 2 n = Q 2 

'. / n 2 (sin 2 n + cos 2 6 n ) = P 2 + 2 

i- e - k = V(P 2 +Q*) 

= y/(AE 2 + AN 2 ) (6.4) 

AE 
= iinT < 65 > 



AN 
cos6 n 

6.43 Where the length and bearing of a line are missing 
The two previous cases provide the required values. 

6.44 Where the bearings of two lines are missing 
(1) // the bearings are equal 

l p sind p + l q sind Q = P 



l p cos d p + l Q cos6 q = Q 



*% = e q = 



Then l p smd + l g sin$ = P 



or 



«* 


+ lg)sin0 


= 


P 






sin 6 


= 


P 

l v + 


la 




cos 6 


= 


Q 

l p + 


U 




tan# 


= 


P 
Q 





(2) // the bearings are adjacent 

/,sin#, + / 2 sin0 2 = P 
/,cos<9, + l 2 cosd 2 = Q 



(6.6) 



(6.7) 



310 



SURVEYING PROBLEMS AND SOLUTIONS 



In Fig.6.8, /,, l 2 , Z 3 , l A , <9 3 ,and 
6 4 are known. 

AC = l s can be found with the 
bearing AC. 

In triangle ABC, 

tan ° _ fa-bKs-.c) 
2 "V S (s - a) 



where 



sinS 



a + b + c 

2 
b sin a 




Fig. 6.8 



From the value of the angles the required bearings can be found. 

Bearing AB = bearing AC - angle a 

Bearing BC = bearing BA - angle B 
(3) // the bearings are not adjacent 




Fig. 6.9 
Assume 6 ' and 6 are missing. 

Graphical solution (Fig. 6.9) 

Plot the part of the survey in which the lengths and bearings are 
known, giving the relative positions of A and D. 

At A draw circle of radius AB - /,; this gives the locus of station 
B. 

At D draw circle of radius DC = l 3 ; this gives the locus of station 
C. 

From A plot length and bearing l 2 6 2 to give line AH. 

At H draw arcs HC X and HC 2 , radius /,, cutting the locus of 
station C at C, and C 2 . At C, and C 2 draw arcs of radius BC - l 2 , 
cutting the other locus at B, and B z - 



TRAVERSE SURVEYS 



311 



Mathematical solution 

Using the graphical solution: 

Find the length and bearing AD. Solve triangle AHD to give HD. 
Solve triangle HC^D to give <f> and /3 and thence obtain the bearings 
of tfC, = bearing AB X , HC 2 = bearing AB Z , C,D and C Z D. 

N.B. There are two possible solutions in all cases (1), (2) and (3), 
and some knowledge of the shape or direction of the lines is required 
to give the required values. 

Alternative solution 

Let /,sin0, + / 3 sin0 3 = P (1) 

/,cos0, + / 3 cos 3 = Q (2) 

i.e. ^sintf, = P - / 3 sin0 3 (3) 

/,cos0, = Q - / 3 cos0 3 (4) 

Squaring (3) and (4) and adding, 

If = P 2 + Q z + \\ - 2/ 3 (P sin 3 + Q cos 3 ) 



P 

Vtf^+Q 2 ) sin $3 + V(P 2 + Q 2 ) C0S ° 3 

= P 2 + Q z +H-l? 
2l 3 y/(P 2 +Q 2 ) 

Referring to Fig. 6.10 




sin a sin0_ + cos a cos0 a = 



P 2 + Q 2 + l\ - I 2 



2l 3 yJ(P 2 +&) 

i.e. cos(0 3 - a) = k 

6 3 - a = cos"'/: 



a = tan 



rP 
Q 



= k (6.8) 



from (3), 



a = cos -1 fc + tan -1 — 
Q 

. P - / 3 sin0_ 
SU10, = 2 § 



(6.9) 
(6.10) 



Example 6.2 The following data relate to a closed traverse ABCD 
in which the bearings of the lines AB and CD are missing. 



312 



SURVEYING PROBLEMS AND SOLUTIONS 



Length 


Bearing 


AB 200 


- 


BC 350 


102° 36 ' 


CD 150 


- 


DA 400 


270° 00' 


Calculate the missing data. 




Method (1) Fig. 6. 11 




s. 





Fig. 6.11 




In triangle ADE, AE = BC = 350 
AD = 400 

By co-ordinates relative to A, 
st.(D) 400 E ON 

st.(E) 350 sinl02°36' = +350 sin77°24' = +341-57 
350 cos 102° 36' = - 350 cos 77° 24' = - 76-34 

tan bearing ED _ ™ - 34 ^ 57 _ «*43 
+ 76-34 76-34 

bearing ED = N37°26'E = 037° 26' 

length ED = 76-34 sec 37° 26' = 96-14 

58-43 cosec37°26' = 96-13 {check) 



In triangle EDC, 



where s 



ED + DC + EC 



I [ (126-93) (73 -07) ) 
A/l(223-07)(23-07)j 



<f>/2 = 53° 19' 
cf> = 106° 38' 



check 126-93 

73-07 

23-07 

s 223-07 



TRAVERSE SURVEYS 313 

DC sin<£ 



sin/8 = 



EC 
150 sin 106° 38' 



200 
/3 = 45° 56' 
Bearing AB 2 = 037° 26' + 45° 56' = 083° 22' 

or AB, = 037° 26' - 45° 56' = 351° 30' 

Bearing DC, = 217° 26' + 106° 38' = 324° 04' 

or DC 2 = 217° 26' - 106°38' = 110° 48' 

.". Bearing CD = 144° 04' 

or 290° 48' 

Method (2) 

200 sin 0, + 350 sin 102° 36' + 150 sin 2 + 400 sin 270° = (1) 
200cos<9, + 350cosl02°36' + 15Ocos0 2 + 400cos270° = (2) 

i.e. 200 sin 0, + 150 sin 2 = -341*57 + 400 = 58*43 (3) 

200 cos 0, + 150 cos 2 = 76*34 + = 76*34 (4) 

.*. 200 sin0, = 58*43 - 150 sin0 2 (5) 

200 cos 0, = 76*34 - 150 cos 2 (6) 

Squaring and adding, 
200 2 = S8-43 2 + 76*34 2 + 150 2 - 2 x 150 (58*43 sin 2 + 76*34 cos 2 ) 

• cq^q • a koa a 58*43 2 + 76*34 2 + 150 2 -200 2 

• • 58*43 sin O + 76*34 cos 0, = 



cos (0 2 - a) 



300 

5S-43 2 + 76*34 2 + 150 2 - 200 2 

300 V(58*43 2 + 76*34 2 ) 
3414*07 + 5827*79 + 22500 - 40 000 



300 V(3414*07 + 5827*79) 

2 - a = -73° 22' 

= 106° 38' or 253° 22' 

. 4 t 58*43 

but tana = 

76*34 

a - 37° 26' 

••• 0, = 144° 04' or 290° 48' 



314 
from (5) 



SURVEYING PROBLEMS AND SOLUTIONS 

58*43- 150 sin 144° 04' 



sin 6. = 



200 



0, = 351° 30' 



or sin#, = 



*, 



58-43 -150 sin 290° 48' 



200 



83° 21' 



6.45 Where two lengths are missing 

Let /,sin0, + / 2 sin0 2 = P 

/,cos#, + / 2 cos0 2 = Q 

(a) The simultaneous equations may be solved to give values for / and 
l 2 regardless of their position. 

(b) If they are adjacent lines the solution of a triangle ADE will give 
the required values (Fig. 6.12), as length AD together with angles a 
and /3 are obtainable* 



Fig. 6.12 




~"X> e * 



(c) If 0, = 6 2 =6 (i.e. the lines are parallel) 

(/,+ / 2 )sin0 = P 

(/,+ / 2 )cos0 = Q 

Squaring and adding, 

</ f +/ a ) a = P 2 + Q Z 

Therefore this is not determinate. 

(d) If /, = l 2 = I and 6, = d 2 = 6 

2/sin0 = P 

P 



/ = 



2 sin<9 



(6.11) 



♦The lines can be adjusted so that the missing values are adjacent. Solution 
(b) can then be applied. 



TRAVERSE SURVEYS 315 

6.46 Where the length of one line and the bearing of another line are 
missing 

I 
Let /,sin0, + / 2 sin0 2 = P 

/,cos0, + / 2 cos0 2 = Q 

where 0, and l 2 are missing. 

Then, as before, 

/, sin0, = P - l 2 sin0 2 (1) 
Z,cos0, = Q - / 2 cos0 2 (2) 

Square and add, 

I 2 = P 2 + Q 2 + l\- 2/ 2 (Psin0 2 + Qcos0 2 ) 
this resolves into a quadratic equation in l 2 . 

I 2 - 2/ 2 (P sin0 2 + Q cos0 2 ) + P 2 + Q 2 - if = (6.12) 
Then from the value of Z 2 , 0, may be obtained from equation (1). 

Example 6.3 Using the data of a closed traverse given below, calcul- 
ate the lengths of the lines BC and CD. 
Line Length (ft) W.C.B. Reduced bearing Latitude Departure 

N 14°31'E +333-0 + 86-2 

N 40° 18' W 
N 12°45'W 

N 05°16'E +298*8 + 27-6 

S 11°48'E -1916-4 +400-4 

(I.C.E.) 

Assuming the co-ordinates of A to be + 1000 E, + 1000 N, from the 
given co-ordinates: 



AB 


344 


014° 31' 


BC 




319° 42' 


CD 




347° 15' 


DE 


300 


005° 16' 


EA 


1958 


168° 12' 



*A 


+ 1000-0 


N* 


+ 1000-0 


&£ AE 


- 400-4 


^ae 


+ 1916-4 


*E 


+ 599-6 


N* 


2916-4 


&E ED 


- 27-6 


^ED 


- 298-8 


*D 


+ 572-0 


K D 


2617-6 


Ea 


+ 1000-0 


K A 


+ 1000-0 


^AB 


+ 86-2 


^AB 


+ 333-0 


E* 


+ 1086-2 


N* 


+ 1333-0 



AE™ - 514-2 AN« n + 1284*6. 



-"BD Jlii ^"BD 



316 



SURVEYING PROBLEMS AND SOLUTIONS 



Bearing BD = tan"'- 514*2/+ 1284'6 
= N21°49'W = 338° 11' 



BC = 319° 42' 

Angle CBD = 18° 29' 

DB = 158° 11' 

DC = 167° 15' 

Angle BDC = 9° 04' 

(fi + D) = 27° 33' 

Length BD = 1284'6/cos21°49' 

= 1383-7 

In triangle BCD, 
DB sin B 



dn 9 e 



(B) 



(D) 



DC = 



sin(B + D) 



BC = 



1383-7 sin 18° 29' 
sin 27° 33' 
= 948-4 ft 
DB sin D 1383-7 sin 9° 04' 



sin(B + D) 



sin 27° 33' 
471 -4 ft 




18° 29' 



Exercises 6(b) (Omitted values) 

8. A clockwise traverse ABCDEA was surveyed with the following 
results: 

AB 331-4 ft EAB 128° 10' 20" BCD 84° 18' 10" 

BC 460-1 ft 

CD 325-7 ft ABC 102° 04' 30" CDE 121° 30' 30" 



TRAVERSE SURVEYS 317 

The angle DEA and the sides DE and EA could not be measured 
direct. Assuming no error in the survey, find the missing lengths and 
their bearings if AB is due north. 
(L.U. Ans. EA = 223-lft, DE = 293-7ft, 308° 10' 20", 232° 06' 50" ) 

9. An open traverse was run from A to E in order to obtain the length 
and bearing of the line AE which could not be measured direct, with 
the following results: 

Line AB BC CD DE 

Length 1025 1087 925 1250 

W.C.B. 261° 41' 9° 06' 282° 22' 71° 30' 
Find by calculation the required information. 

(L.U. Ans. 1901; 342° 51') 

10. The following measurements were obtained when surveying a 
closed traverse ABCDEA: 



Line 


EA 


AB 


BC 




Length (ft) 


793-7 


1512-1 


863-7 






DEA 


EAB 


ABC 


BCD 


Included angles 


93° 14' 


122° 36' 


131° 42' 


95° 43' 



It is not possible to occupy D, but it could be observed from both 
C and E. 

Calculate the angle CDE, and the lengths CD and DE, taking DE 
as the datum, and assuming all observations to be correct. 

(L.U. Ans. 96° 45'; 1848-0 ft, 1501-6 ft) 

11. In a traverse ABCDEFG, the line BA is taken as the reference 
meridian. The latitudes and departures of the sides AB, BC, CD, DE 
and EF are: 

Line AB BC CD DE EF 

Latitude - 1190*0 - 565*3 + 590'5 + 606*9 + 1097*2 
Departure + 736*4 + 796*8 - 468*0 + 370*4 

If the bearing of FG is N 75°47'W and its length is 896'Oft, find 
the length and bearing of GA. 

(L.U. Ans. 947-6 ft; S 36°45'W) 

6.5 The Adjustment of Closed Traverses 

(a) Traverses connecting two known points. 

(b) Traverses which return to their starting point. 

6.51 Where the start and finish of a traverse are fixed 

The length and bearing of the line joining these points are known 
and must be in agreement with the length and bearing of the closing 
line of the traverse. 



318 



SURVEYING PROBLEMS AND SOLUTIONS 



Where the traverse is not orientated to the fixed line an angle of 
swing (a) has to be applied. 

Where there is discrepancy between the closing lengths, a scale 
factor k must be applied to all the traverse lengths. 



k = 



length between the fixed points 
closing length of traverse 




Fig. 6.14 

In Fig. 6.14, the traverse is turned through angle a so that traverse 
ABCD becomes AB^D and AD } is orientated on to line XY. The 

XY 

scale factor k ^~rf\ must be applied to the traverse lines. 

Traverses are often orientated originally on their first line. Co- 
ordinates are then computed, and from these the length and bearing of 
the closing line (AD). The latter is then compared with the length and 
bearing XY. 

Co-ordinates of the traverse can now be adjusted by either 

(1) recomputing the traverse by adjusting the bearings by the 
angle a and the length by multiplying by k, or 

(2) transposing the co-ordinates by changing the grid (Eqs. 
3.33/3.34) and also applying the scale factor k, or 

(3) applying one of the following adjustment methods, e.g. 
Bowditch. 

N.B. The factor k can be a compounded value involving: 

(a) traverse error, 

(b) local scale factor — (ground distance to national grid), 

(c) change of units, e.g. feet to metres. 

Example 6.4. A traverse XaY is made between two survey stations 
X (E 1000 N 1000) and Y (E 1424*5 N754'9). 



TRAVERSE SURVEYS 319 

Based upon an assumed meridian, the following partial co-ordinates 
are computed: 

AE AN 

Xa 69-5 - 393*9 

aY 199-3 - 17-4 

Adjust the traverse so that the co-ordinates conform to the fixed 
stations X and Y. 

Ans. E N 

X 1000-0 1000-0 

Y 1424-5 754-9 



AE + 424*5 AN - 245*1 

Bearing of control line XY = tan"' 424*5/- 245*1 

= S 60° OO'E i.e. 120° 00' 
Length of control line XY = 424*5/sin60° 

= 490*2. 
From the partial co-ordinates, 

AE XY = 69*5 + 199*3 - + 268*8 
AN^ y = -393*9- 17-4 = -411*3 
Bearing of traverse line XY = tan -1 + 268*8/- 411*3 

= S33°10 , E i.e. 146° 50' 
Length of traverse line XY = 411*3/cos33° 10' 

= 491*4. 
Angle of swing a = traverse bearing XY - fixed bearing XY 
= 146° 50' - 120° 00' 
= + 26° 50' 

Scale factor k = fixed length/traverse length 
= 490*2/491*4 
= 0*99756 

Using Eqs.(3.33) and (3.34), 

AE' = + mAE - nAN 

AN' = mAN + «AE 
m = kcosa = 0*99756 cos 26° 50' = 0*89014 
n = ksina = 0*99756 sin 26° 50' = 0*45030 



320 



SURVEYING PROBLEMS AND SOLUTIONS 



Line 

Xa 

aY 



Ae An 

+ 69-5 -393-9 
+ 199-3 - 17-4 



mAE nAN 

+ 61-86 -177-37 
+ 177-40- 7-84 



Total co-ordinates 
X 
a 
Y 



Ae' 

+ 239-23 
+ 185-24 
+ 424-47 

E 
1000-0 
1239-2 
1424-5 



+ 

mAN nAE 

-350-63 +31-30 

- 15-48 +89-74 

N 

1000-0 

680-7 

754-9 



An' 

-319-33 
+ 74-26 
-245-07 



Example 6.5 If in the previous example the co-ordinates of Y are 

E 1266-9 N 589-1, 

then the bearing of XY = tan -1 + 266*9/ - 410-9 

= S33°00'E i.e. 147° 00' 
length XY = 410-9/cos33° 
- 490-0 
angle of swing a = 146° 50' - 147° 00' 
= -0°10' 
a fad = -0-002 91 
Scale factor k = 490*0/491-4 
= 0-99716 
Using Eqs. (3.35) and (3.36), 

AE' = fc[AE - ANa] 
AN' = fc[AN + AEa] 



Line 

Xa 

aY 



Ae An 

+ 69-5 -393-9 
+ 199-3 - 17-4 



ANa AEa 

+ 1-15 -0-20 
+ 0-05 -0-58 



Ae - ANa An + AEa 

+ 68-35 -394-10 
+ 199-25 - 17-98 



Ae' An' 

+ 68-16 -392-98 
+ 198-68 - 17-93 



2 +266-84 -410-91 



Example 6.6 An underground traverse between two wires in shafts A 
and D based on an assumed meridian gives the following partial co- 
ordinates: 

AE(ft) AN (ft) 

AB 263-516 

BC + 523-684 + 21*743 

CD + 36-862 +421*827 



If the grid co-ordinates of the wires are: 



TRAVERSE SURVEYS 



321 



A 

D 



E (metres) 
552361-63 
552 532-50 



N (metres) 
441 372-48 
441428-18 



Transform the underground partials into grid co-ordinates. 

Arts. 

Bearing of traverse line AD = tan" 1 + 560*546/ + 180-054 - N72°ll'33"E 

Length of traverse line AD = 560*546 /sin 72° 11 '33" = 588*755 ft 

Bearing of grid line AD = tan" 1 170-87/55*70 = N71°56'45"E 

Length of grid line AD = 170*87/sin71° 56'45" = 179*713 m 

Angle of swing a = 72°11'33" - 71°56'45" = +0°14'48" 

a rad = 0*004 31. 
Scale factor k = 179*713/588*755 = 0*30523 
Using Eqs.(3.35) and (3.36), 

AE' = fc[AE - ANa] 

AN' = fc[AN + AEa] 



Line 


AE 


AN 


ANa AEa 


AE - ANa AN + AEa 


AE' AN' 


AB 


o-o 


- 263-516 


- 1-136 


+ 1-136 -263-516 


+ 0-35 - 80-43 


BC 


+ 523-684 


+ 21-743 


+ 0-094 +2-257 


+ 523-590 + 24-000 


+ 159-82 + 7-33 


CD 


+ 36-862 


+ 421-827 


+ 1-818 + 0-159 


+ 35-044 +421-986 


+ 10-70 + 128-80 






X +170-87 + 55-70 






E N 






A 552 361 -63 m 441 372-48 m 






B 552 361-98 m 441 292-05 m 






C 552 521 -80 m 441 299-38 m 








D 552 532-50 m 


441 428-18 m 





Example 6.7 Fig. 6.15 shows a short 'dial' traverse connecting two 
theodolite lines in a mine survey. The co-ordinates of T.M.64 are 
E 45603-1 ft N 35 709-9 ft and of T.M.86 E 46 163*6 ft N 35 411*8 ft. Cal- 
culate the co-ordinates of the traverse and adjust to close on T.M.86. 

(N.R.C.T.) 



241° 26' 



216° 57' 



251° 06' 




Fig. 6.15 



T.M.87 



322 



SURVEYING PROBLEMS AND SOLUTIONS 



This is a subsidiary survey carried out with a 1 minute instrument 
(miners' dial) and the method of adjustment should be as simple as 
possible. 

Bearing T.M. 63- T.M. 64 048° 19' 
+ angle 241° 26' 







289° 45' 


i 


Adjusted Bearings 








-180° 








Bearing T.M. 64 -(1) 




109° 45' 


+ 01' 


109° 46' S70°14' 


E 


+ angle 




216° 57' 

326° 42' 
- 180° 








Bearing 1-2 




146° 42' 


+ 02' 


146° 44' S33°16 


E 


+ angle 




110° 41' 
257° 23' 
- 180° 








Bearing 2 -T.M. 86 




077° 23' 


+ 03' 


077° 26' N77°26 


E 


+ angle 




251° 06' 
328° 29' 
- 180° 








Bearing T.M. 86 - T.M 


.87 


148° 29' 


+ 04' 


148° 33' S31°27 


E 


Fixed bearing T.M. 86- 


T.M.87148°33' 









.-. Error 04' 

Horizontal length T.M. 64-1(1 in 6 = 9° 28') = 286*1 cos 9° 28' = 282'2 ft 
Co-ordinates 

Ae Se Ae' 



Line 


Length 


64-1 


282-2 


1-2 


273-2 


2-86 


146-4 



Bearing 
S 70°14'E 
S 33°16'E 
N 77°26'E 



+ 265-6 - 0-5 + 265-1 
+ 153-2 - 0-5 + 152-7 
+ 142-9 - 0-2 + 142-7 



+ 561-7 - 1-2 



An 


Sn 


An' 


- 95-4 


-0-5 


- 95-9 


- 233-5 


- 0-4 


- 233-9 


+ 31-9 


- 0-2 


+ 31-7 


+ 31-9 


- 1-1 




- 328-9 






- 297-0 







From the theodolite station co-ordinates, 

E N 

T.M. 64 45603-1 35 709*9 

T.M. 86 46163-6 35 411-8 



AE + 560-5 AN - 298-1 
'• Error in traverse = E + 1*2, N + 1*1 



TRAVERSE SURVEYS 323 



Applying Bowditch's method (see p. 330), 
a-, 1*2 x length 1*2 x / 







2 length 


701-8 


- 1/1 X 


1U 




SN 


1-1 x / 
2/ 


1-1 X 
701-8 


- = 1-57 x 


10~ 3 


Total 


co-ordinates (adjusted) 














E 


N 








T.M.64 


45603-1 


35709-9 








1 


45868-2 


35614-0 








2 


46020-9 


35380-1 





X /. 



86 46163-6 35411-8 

6.52 Traverses which return to their starting point 

The closing error may be expressed as either (a) the length and 
bearing of the closing line or (b) the errors in latitude and departure. 

To make the traverse consistent, the error must be distributed and 
this can be done by adjusting either 

(a) the lengths only, without altering the bearings or 

(b) the length and bearing of each line by adjusting the co-ordinates. 

6.53 Adjusting the lengths without altering the bearings 

Where all the angles in a closed traverse have been measured, the 
closing angular error may be distributed either (a) equally or (b) by 
weight inversely proportional to the square of the probable error. 

It may therefore be assumed that the most probable values for the 
bearings have been obtained and that any subsequent error relates to 
the lengths, i.e. a similar figure should be obtained. 

Three methods are proposed: 

(1) Scale factor axis method. 

(2) xy method (Ormsby method). 

(3) Crandal's method. 

In the following description of these methods, to simplify the solu- 
tion, the normal co-ordinate notation will be altered as follows: partial 
departure AE becomes d with error in departure Sd. The sum of the 
errors in departure XSd becomes Ad. Similarly, partial latitude AN 
becomes / with error in latitude 81. The sum of the errors in latitude 
25/ becomes A/. 

(1) Scale factor axis method (after R.E. Middleton and Q. Chadwick). 
This follows the principles proposed for traverses closed on fixed 
points. 



324 SURVEYING PROBLEMS AND SOLUTIONS 

Graphical solution (Fig. 6. 16) 

The traverse is plotted and the closing error obtained. 

It is intended that this error produced should divide the figure into 
two approximately equal parts. To decide on the position of this line, 
the closing error bearing is drawn through each station until the above 
condition is obtained. 



(a) Original plot with 
lines drawn parade 
to closing error A A 




(b) Adjusted figure \^/ 



Fig. 6.16 Scale factor axis method 



The lines above XD need to be reduced by a scale factor so as to 
finish at D 2 midway between D and D,. The lines below XD need 
to be enlarged by a scale factor so as to finish at D 2 . 



TRAVERSE SURVEYS 



325 



Example 6.8 (Fig. 6.17) 




Fig. 6.17 Graphical adjustment 

(1) The traverse is plotted as ABCDEA^ -closing error AA y . 

(2) The closing error is transferred to station D, so that the traverse 
now plots as D^E^ABCD. D^D 2 is produced to cut the traverse 
into two parts at X on line AB. 

(3) From X, rays are drawn through B, C, D, £, and A. 

(4) From 0, midway between D, and D, lines are drawn parallel to 
D,/},, giving 0E 2 , and parallel to DC, giving 0C 2 . Lines paral- 
lel to BC and EA^ through C 2 and E 2 give B z and A 2 . 

The figure A 2 B 2 C 2 0E 2 A 2 is the adjusted shape with the bear- 
ings of the lines unaltered, i.e. 

OX 
Lines XB, BC and CD are reduced in length in the ratio jrzs 

OX 
Lines XA, AE and ED y are enlarged in length in the ratio pr~- 

(5) If the traverse is to be plotted relative to the original station A, 
then all the new stations will require adjustment in length and bear- 
ing A Z A, giving B 3 , C 3 , D 3 , E 3 . 

The final figure is AB 3 C 3 D 3 E A. 



326 



SURVEYING PROBLEMS AND SOLUTIONS 



The graphical solution can now 
be expressed mathematically as 
follows (after R.E. Middleton and 
O. Chadwick). 

In the figure XBCD, Fig. 6.18, 
the lengths need to be reduced by 
a scale factor 

1 XD 

As the co-ordinates are dependent 
on their respective lengths, the par- 
tial co-ordinates can be multiplied 
by the factor k v 

Let the partial departure be d 
and the partial latitude /, the error 
in d be 8d and the error in / be 81. 
Expressing this as a correction, 

8d = d -dk, = d(l - fc,) 

XD-XD 




d- 



XD 
D,D 



= d^- 
XD 

_ partial departure x terror 
length of axis of error 

Thus for all lines above the line XD, 

dd = Ad 

traverse error (£),D) 



where 



Also 



X = 



2 x axis of error (XD) 

81 = XI 



(6.13) 



(6.14) 



(6.15) 



Similarly in figure XD^E^A the lengths need to be enlarged by a 
scale factor 



8d 



XD 2 

~xb~. 



= d(k z -\) 

= d ( XD * - XD ) 



= d. 



XD, 

D£, 
XD, 



TRAVERSE SURVEYS 



327 



Thus similarly for all lines below line XD. 
8d = fxd 

traverse error (DDi) 



(6.16) 



where 



M 



2 x axis of error (XD t ) 
Also 81 = fil (6.17) 

By comparison with the traverse lines DD, is small and thus 

XD, ~ XD 21 XD 2 
A ~ /z 
Summary 

Sd = Ad 
5/ = A/ 

The sign of the correction depends on the position of the line, i.e. 
whether the line needs to be reduced or enlarged. 
N.B. A special case needs to be dealt with, viz. the line AB inter- 
sected by line DD, produced to X. 

XB must be reduced 
AX must be enlarged. 
XB 



8d 



XB 



8d 



AX 



AX 

= + V d AB X^g 



8d 



l AB 



= fid 



l AB 



Also 



°MB = V-IaB 



AX-XB 
AB 

AX- XB 
AB 



(6.18) 
(6.19) 



(2) xy (Ormsby) method (Fig. 6.19) 
If any line is varied in length by 
a fractional value then the partial 
co-ordinates will be varied in the 
same proportion without altering 
the bearing, i.e. 

8s _ 8d_ _ 8l_ 

s ~ d ~ T 

Let (a) the lines in the NE/SW quad- 
rants be altered by a factor 
x and the lines in the NW/SE 
quadrants by a factor y, 




Fig. 6. 19 



328 



SURVEYING PROBLEMS AND SOLUTIONS 



(b) the sign of all the terms in summation of the partial co-ordin- 
ates in one of the equations (say Eq.6.20) be the same as the 
sign of the greater closing error, 

(c) the sign in the other equation (say Eq.6.21) be made consis- 
tent with the figure, to bring the corrections back to the same 
bearing (Fig. 6. 20). 





r~ 


— "71 




i 
i 


\j- 




i 
i 


s + 






Ad>AI 




+ 








a/ 










i j 


/ \Al 










I / 




/ + 




+ 











i.e. 





Fig. 6.20 


+ Ad = 


xd x + yd z + xd z + yd 4 


+ Ad = 


x(d, + d 3 ) + y(d 2 + d 4 ) 



(6.20) 



+ A/ = x/, - y/ 2 + xl 3 - y/ 4 . 
i.e. + A/ = x(/, + / 3 ) - y(/ 2 + / 4 ) 
where the partial co-ordinates are d, /,; d 2 l 2 , etc. 



(6.21) 



The solution of these simultaneous equations gives values for x 
and y which are then applied to each value in turn to give the correc- 
ted values of the partial co-ordinates. 

(3) CrandaVs method by applying the principle of least squares, 
Fig. 6.21. 

Let the length of each side be varied by a fraction x of its lengths, 
then if / and d be the partial latitude and departure of the line AB 
of bearing 6, they also will be varied by xl and xd respectively. 

If the probable error in length is assumed to be proportional to V s ' 
then the weight to be applied to each line will be 1/s, i.e. 



TRAVERSE SURVEYS 



329 



wt oc 



1 



(probable error) 2 
P.E. OC y/J 

wt oc — 
s 

(The value of x is thus depen- 
dent on the length of the line.) 

By the theory of 'Least Squares', 
the sum of the weighted residual 
errors should be a minimum. 



i.e. 2 



x*s* 



= 2 



x s = minimum. 




Let 



Fig. 6.21 

Ixl = 25/ = A/, total error in latitude. 

?Lxd = 18d = Ad, total error in departure. 

Then differentiating these equations and equating them to zero 
will give the minima values, i.e. 



xs 8x = 
ISx = 
d8x = 



(1) 
(2) 
(3) 



The differentiated equations (2) and (3), being conditional equa- 
tions, should now be multiplied by factors (correlatives) -&, and -k. 
(Reference: Rainsford Survey Adjustments and Least Squares). 

Adding all three equations together and equating the coefficients 
of each 8x to zero, we have 

8x(x, s, - fc, /, - k 2 d,) = 
8x(x 2 s 2 - /c, l 2 - k 2 d 2 ) = etc. 



Thus 



fc,/, + fcjjd, 



x, = 



fc,/ 2 + k 2 d 2 



etc. 



Substituting the values of x into the original equations (2) and 
'3), we have 



h f * f/| +/c2f M + / / M2 +k 2 d 2 \ 



+ ... = A/ 



330 SURVEYING PROBLEMS AND SOLUTIONS 

i.e. fc,£— +k 2 2— =M (6.22) 

aiso d, (*■'■ * M.J + da ^ilhh^ + ... . Ad 

i.e. *, 2 — + * 2 2- = Ad (6.23) 

Solving Eqs.(6.22) and (6.23), we obtain values for fc, and k 2 . 
The corrections to the partial co-ordinates then become 

81 - k ^ / ,dl 

(6.25) 



s 2 s etc. (6.24) 



8d, = fc, Ai + ^ 2 i etc 



s ' s 

It should be noted that a check on the equations is given by 

_5__L_JLl_ _ UMi + MI _A_ 

fe AA A; ^1 d '^ ! Zf + ^^ dl 

1 s + 2 s 

i.e. no change in bearing. 

The assumption that the probable error in length is proportional to 
\Js applies to compensating errors. It has been shown that, where the 
accuracy of the linear measurement decreases, the probable error in 
length becomes proportional to the length itself, i.e. 

P.E. oc s 

1 
i.e. wt oc — r 

s 2 

The effect of this on the foregoing equations is to remove the 
factor s from them, i.e. 

81, = /c ] ^+^/ 1 d 1 etc (6.26) 

Sd, = k^d y + k 2 d* etc (6.27) 

6.54 Adjustment to the length and bearing 

Three methods are compared: 

(1) Bowditch, (2) Transit (Wilson's method), (3) Smirnoff. 

(1) The Bowditch method (Fig. 6.22). This method is more widely 
used than any other because of its simplicity. It was originally devi- 
sed for the adjustment of compass traverses. 



TRAVERSE SURVEYS 



331 



Bowditch assumed that (a) the 
linear errors were compensating and 
thus the probable error (P.E.) was 
proportional to the square root of 
the distance s, i.e. 

P.E. oc y^- 

and (b) the angular error 86 in the 
d would produce an equal displace- 
ment B, B 2 at right angles to the 
line AB. 

A resultant AB 2 is thus develo- 
ped with the total probable error. 

BB Z = Vfi.S 2 + B,Bi 
= y/2B^B 

(BtB = B,B 2 ) 




Fig. 6.22 



also = y/Sl 2 + Sd 2 

where 81 and 8d are the corrections to the partial co-ordinates. 
The weight, as before, becomes 1/s. 
By the theory of least squares 



s («' + "*) ■ 



a minimum 



The conditional equations are: 

181 = A/ (1) 

28d = Ad (2) 

As in the previous method, using correlatives, differentiation of 
each equation gives: 

^ 81 8(8Q 8d 8(8d) 

2* + = 

s s 

28(81) = 

2 8(8d) = 

Multiplying the last two equations by the correlatives -fc, and -k z 
respectively, adding the equations and equating the coefficients of 
5(50 and 5(5d) to zero we have: 



5(5/) 



[H ■ 







i.e. 5/, = s,*, dl z = s 2 k, etc. 



5(5d)|— ~K\ = i.e. 5d, = 5,^ 
Substituting the values into equations (1) and (2), 



8d z = s z k. 



etc. 



332 SURVEYING PROBLEMS AND SOLUTIONS 



A/ 
fc, 2 s = A/ i.e. k x = ■=- 



k z 2s = Ad i.e. k 2 = 
The corrections to the latitudes 67, = s 



2s 

Ad 

Is 

A/_ 
2s 



(6.28) 



Ad 
sL = s, =— - etc. 
^ 2s 

Ad 
to the departures od, = s. -=— 

2s 

Ad 
od, = s, ■=— etc. 

2s 

i.e. Correction to the partial co-ordinate = total correction 

length of corresponding side 

x ______________ _ ___ __ ______ 

total length of traverse 

The effect at a station is that the resultant BB 2 will be equal to 

the closing error x s/2s and parallel to the bearing of the closing 

error, 

-1 -d 
i.e. through an angle a = tan — 

, Sd 

= tan" 1 — 

5/ 

The total movement of each station is therefore parallel to the 
closing error and equal to 

2 (lengths up to that point) , . 

x closing error. 

total length of traverse 

The correction can thus be applied either graphically in the manner 
originally intended or mathematically to the co-ordinates. 

Jameson points out that the bearings of all the lines are altered 
unless they lie in the direction of the closing error and that the maxi- 
mum alteration in the bearing occurs when the line is at right angles 
to the closing bearing, when it becomes 

2s x closing error closing error 



86 



rad 



2s 



The closing error expressed as a fraction of the length of the tra- 
verse may vary from 1/1000 to 1/10000, so taking the maximum error 
as 1/1000 



TRAVERSE SURVEYS 



333 



86" = 2 ° 6265 = 206" 
1000 

= 03' 26" 

a value far in excess of any theodolite station error. A change of bea- 
ring of 20" represents 1/10000 and this would be excessive even 
using a 20" theodolite. 

Graphical Solution by the Bowditch Method (Fig. 6.23) 





Fig. 6.23 

(1) Plot the survey and obtain the closing error AA E . 

(2) Draw a line representing the length of each line of the traverse 
to any convenient scale. 

(3) At A E draw a perpendicular A E A^ , equal to the closing error 
and to the same scale as the plan. 

(4) Join AA X , forming a triangle AA X A E , and then through B, C 
and D similarly draw perpendiculars to cut the line AA y at B„ 
C, and D, . 

(5) Draw a line through each station parallel to the closing error 



334 SURVEYING PROBLEMS AND SOLUTIONS 

and plot lines equal to BB U CC X and DD X , giving the new 
figure Afi,C,D,A . 
(2) The Transit or Wilson method. This is an empirical method 
which can only be justified on the basis that (a) it is simple to ope- 
rate, (b) it has generally less effect on the bearings than the Bowditch 
method. 

It can be stated as: 
the correction to the partial co-ordinate 

., .. f ,. A closing error in the co-ordinate 

= the partial co-ordinate x 



laie x — 

Z partial co-ordinates 

(ignoring the signs) 


(6.29) 




(6.30) 


8d < - d ' Td 


(6.31) 



(3) The Smirnoff method. The partial latitude / of a line length s 
and bearing 6 is given by 

/ = s COS0. 

If the two variables s and cos0 are subjected to errors of 8s and 
8(cosd} respectively, then 

1 + 81 = (s + 8s)[cos0 + S(cos0)] 

Subtracting the value of / from each side and neglecting the small 
value 8s S(cos0), gives 

81 = 8s cos# + s 8(cosd) 
Dividing both sides by /, 

— _ 8s cos e s _S(cos0) 
/ s cos0 s cos 6 

i.e. 81 8s 8(cosd) 

j = — + , ' (6.32) 

/ s cos# 

i.e. the relative accuracy in latitude 

= the relative accuracy in distance + 
the relative accuracy in the cosine of the bearing 

Thus, in a traverse of n lines, 

M 1 8s * j. i g ( cos fl) 

s, cos 6/, 

Sl 2 = I — + l ——— — etc. 

s 2 - cosu 2 



TRAVERSE SURVEYS 335 

25/ = A/ (total error in latitude) 

v 8s SCcosfl, ) S(cos0 2 ) 5(cos0 n ) 

— 2*1 + I. — + / 2 — — + ... l n 

s cos0, " cos0 2 cos0 n 



Similarly, as d = s sin 0, 

— = £f S(sinfl) 
d s sin 



(6.33) 



18d = Ad = 2d«f + d , S(5in ^ ) + w g(sin ^> + .. <* 5 < sin ^ 
s sin0, 2 sin0 2 n sin0 n 

where the linear relative accuracy 8s/ s is considered constant for all 
lines. 



From the above equations, 



S(cos0)l 



8s _ 1_[ 

s ~ 2/1 

^ 1>-Sd«l (6.35) 

s zdl sin0 J 

The value of 8s/ s should closely approximate to the actual acc- 
uracy in linear measurement attained if the traverse consists of a large 
number of lines, but in short traverses there may be quite a large disc- 
repency. In such cases the ratio shows the accuracy attained as it 
affects the closing error. 

The ratio is first worked out separately for latitude and departure 
from Eqs. 6.34/6.35 and these allow subsequent corrections to be 
applied as in Eqs. 6.32/6.33. 

N.B. The precision ratios for cosine and sine of the bearings are 
obtained by extraction from trigonometrical tables. Special attention 
is necessary when values of 0° or 90° are involved as the trigonomet- 
rical values of oo will be obtained. The traverse containing such bear- 
ings may be rotated before adjustment and then re-orientated to the 
original bearings. 

To calculate the precision ratio for cos 0, 

Let = 60° + 6" 

cos0 = 0-5 

8 (cos 0) difference/ 6" = 0-000025 

S(cos0) _ 0-000025 1 

cos0 ~ 0^5 = 20000 

The ratio is therefore proportional to the angular accuracy. 

Where a negative ratio 8s/ s is obtained it implies that the angular 
precision has been over-estimated. 



336 



SURVEYING PROBLEMS AND SOLUTIONS 



As the precision of the angular values increases relative to the 
linear values, the precision ratios of the former reduce and become 
negligible. 

8s = Al 
s 11 



Thus 



and, substituting this into the partial latitude equation, 

A/ 



81, = Z, 



21 



Similarly, 



sd, - d, Td 



These equations thus reduce to method of adjustment (2), the 
Transit Rule. 
6.55 Comparison of methods of adjustment 

Example 6.8 



Line 

AB 
BC 
CD 
DE 
EF 
FA 



I 



Bearing 6 


Length s 


+ 


- 


+ 


- 


045° 00' 


514-63 


363-898 




363-898 




090° 00' 


341-36 


341-360 




0-0 




180° 00' 


324-15 


0-0 






324-150 


210° 00' 


462-37 




231-185 




400-420 


300° 00' 


386-44 




334-667 


193-220 




320° 16' 


217-42 




138-978 


167-202 




2s 


2246-37 


705-258 
704-830 


704-830 1 


724-320 


724-570 
724-320 




Ad 


+ 0-428 




A/ 


-0-250 



Assuming the co-ordinates of A (0,0) 
Total Co-ordinates 



D 

0-0 
+ 363-898 
+ 705-258 
+ 705-258 
+ 474-073 
+ 139-406 
+ 0-428 
Bearing of closing line AA, = tan-' +0-428/-0-250 

= S59°43'F 
Length of closing line A A, = 0-428 cosec59°43' 

= 0-496 



A 
B 
C 
D 
E 
F 
A, 



L 

0-0 
+ 363-898 
+ 363-898 
+ 39-748 
-360-672 
-167-452 
- 0-250 



TRAVERSE SURVEYS 



337 



(1) 'Axis' scale factor method (Figs. 6.24 and 6.25) 

400r 



300 



200 




700 800 



-100- 



•200- 



-300- 



-400 



Fig. 6.24 







i 


Lines above axis 




\D 


! 


require reduction 


Lines below axis 




\io«>2> 




require enlargement / 




/ ^*\ 


^i 


'E 


E, 




^Axis 




Fig. 


6.25 





Scaled values from plotting (station D is chosen to contain the closing 
error as the axis approximately bisects the figure): 

DX = 487 
AX ' = 406 
XB = 108 

From Eqs. 6.14/6.15, 8d = \d 
81 = \l 

AA X 0-496 



where 



A = 



8d 



2 x Axis(DX) 2x487 

AX - XB 



= 5-1 x 10" 4 



AB = 



AB 



x Xd AB 



406 - 108 c.-i 1n -4 o*o o 

= — ■=—: x + 5*1 x 10 x 363-9 

514 



= +0-105 



338 



SURVEYING PROBLEMS AND SOLUTIONS 



(AX requires enlarging, AX > XB) 





8d BC = 


-5-1 x 10" 4 x +341-4 


\ 


= -0-174 


SdcD 




= 0-0 


8d DE = +5-1 x 10 -4 x -231-2 




= -0-118 


8d EF = +5-1 x 10 -4 x -334-7 




= -0-171 


8d FA = 

81ab = 
$Ibc 


f5-l x 10" 4 x -139-C 




= -0- 


070 

105 



. s.i in -4 406 - 
f5-l x 10 X r—- A 

514 


108 x +363-9 - +0- 
= 0- 


8l CD = -5-1 x 10" 4 x -324-2 




= +0-165 


8l DE = +5-1 x 10" 4 x -400-4 


[ 


= -0-204 


8l EF = +5-1 x 10" 4 x +193-2 


_ 


= +0-098 


8l FA = +5-1 x 10 -4 x +167-2 




= +0-086 


Co-ordinates 








d 


8d 


d n 


la 


81 




AB 


+ 363-898 


+0-105 


+ 364-063 


+ 363-898 


+ 0-105 


+ 364-063 


BC 


+ 341-360 


-0-174 


+ 341-186 


0-0 


0-0 


0-0 


CD 


0-0 


0-0 


0-0 


-324-150 


+0-165 


-323-985 


DE 


-231-185 


-0-118 


-231-303 


-400-420 


-0-204 


-400-624 


EF 


-334-667 


-0-171 


-334-838 


+ 193-220 


+0-098 


+ 193-318 


FA 


-138-978 


-0-070 


-139-048 


+ 167-202 


+ 0-086 


+ 167-288 


-0-533 




+0-454 


+0-105 




-0-204 


Ad -0-428 


M 


+0-250 


(2) Ormsby's xy method (Fig. 6.26) 








Bearing 


Term 


d 


8d 


/ 


81 


AB 


045° 00" 


X 


+ 363-898 


-0-063 


+363-898 


-0-063 


BC 


090° 00' 


X 


+ 341-360 


-0-060 


0-0 


0-0 


CD 


180° 00' 


y 


0-0 


0-0 


-324-150 


+0-181 


DE 


210° 00' 


X 


-231-185 


-0-040 


-400-420 


-0-069 


EF 


300° 00' 


y 


-334-667 


-0-187 


+ 193-220 


+ 0-108 


FA 


320° 16' 


y 


-138-978 


-0-078 


+ 167-202 


+0-093 


Ad 


-0-428 


+0-382 
-0-132 












A 


I +0-250 



TRAVERSE SURVEYS 



339 



N.B. Term x is assumed to be 0° -» 90° inclusive 

y is assumed to be 90°-* 180° 
As the error in departure is greater, the equation takes the sign of the 
correction, i.e. 

-364 x -341 x - Oy -231 x -335y -139y = -0-428 (1) 

(AB) (SC) (CD)(DE) (£F) (FA) 
Adjusting the latitude values and signs to make them consistent, 

-364 x + Ox + 324y -400 x +193y + 167y = +0-250 (2) 




Fig. 6.26 

Simplifying the equations, 

-936x -474y = -0-428 (1) 

-764* +684y = +0-250 (2) 

Solving the equations simultaneously, 

x = +1-74 x HT 4 
y = +5-59 x 10~ 4 

The values of x and y are now applied to each term to give corrections 
as above. 



(3) CrandaVs method (least squares) 

(a) Probable error oc length s 
Using equations, 

k,2ld + k 2 ^d z 
k,Xl 2 + k 2 lid 



Ad 

A/ 



340 



SURVEYING PROBLEMS AND SOLUTIONS 



AB 
EC 
CD 
DE 
EF 
FA 



d 
+ 363-898 
+ 341-360 

0-0 

-231-185 

-334-667 

-138-978 

Ad -0-428 



* 
+ 363-898 

0-0 
-324-150 
-400-420 
+ 193-220 
+ 167-202 
M +0-250 



+ 132421-8 

116526-6 f 

0-0 

53 446-5 

112002-0 

19 314-9 

433 711-8 

(Id 2 ) 



Id 

+ 132 421-8 
0-0 
0-0 

+ 92571-1 

- 64664-4 

- 23237-4 

+ 224 992-9 

- 87901-8 
+ 137091-1 



(2/d) 

137091-U,+ 433711-8fc 2 = -0-428 
463 121-7 fc,+ 137 091-1 k 2 = +0-250 



Solving simultaneously, 

fc, - +9-1768 x 10~ 7 
k 2 = -1-2769 x 10~ 6 

Substituting in the equations 

/c, /, d, + k 2 d 2 = 5d, 
k,/ t + fc 2 /,d, = S/, 

gives the correction for each partial co-ordinate: 



+ 132421-8 

0-0 

105 073-2 

160 336-2 

37 334-0 

27956-5 

463121-7 



(1) 
(2) 





kyld 


k 2 d z 


Sd 


M 2 


k 2 ld 


81 


AB 


+ 0-122 


-0-169 


-0-047 


+ 0-122 


-0-169 


-0-047 


EC 


0-0 


-0-149 


-0-149 


0-0 


0-0 


0-0 


CD 


0-0 


0-0 


0-0 


+ 0-096 


0-0 


+ 0-096 


DE 


+ 0-085 


-0-068 


+ 0-017 


+ 0-147 


-0-118 


+0-029 


EF 


-0-059 


-0-143 


-0-202 


+ 0-034 


+ 0-083 


+0-117 


FA 


-0-021 


-0-026 


-0-047 


+0-026 


f 0-029 


+ 0-055 




+ 0-207 


-0-555 


-0-445 


+ 0-425 


- 0-287 


+0-297 




-0-080 


+0-127 
-0-428 


+ 0-017 


-0-175 
+ 0-250 


+ 0-112 


-0-047 




+ 0-127 


-0-428 


-0-175 


+ 0-250 



TRAVERSE SURVEYS 



341 



(b) Probable error oc -y/ s 



AB 
BC 
CD 
DB 
EF 
FA 



s 
514-63 
341-36 
324-15 
462-37 
386-44 
217-42 



1/s 
0-001 945 
0-002929 
0-003 085 
0-002 163 
0-002 588 
0-004 599 



d7s 
257-295 
341-306 

0-0 

115-605 

289-861 

88-292 



Id/s 
+ 257-295 

0-0 

0-0 
+ 200-231 
-167-351 
-106-869 



17 s 
257-295 

0-0 
324-151 
346-807 

96-620 
128-572 



2d 2 /s 1092-359 



+ 457-526 S/ 2 /s 1153-445 
-274-220 



Xld/s +183-306 



Using equations 



s s 



k, 2— + k 
s 



s ld 



= Ad 



M 



and substituting values gives 

+ 183-306 k, + 1092-359 k 2 = 
+ 1153-445 k, + 183-306 k 2 = 

Solving simultaneously, 

*, = +2-867 x 10~ 4 
k 2 = -4-398 x 10~ 5 
Substituting values into equations 



-0-428 
+ 0-250 






= 8d, 

8d 
-0-039 
-0-150 

0-0 
+ 0-006 
-0-175 
-0-070 
-0-434 

+0-006 

+0-052 -0-428 -0-428 





/c, Id/s 


k 2 d z /s 


AB 


+ 0-074 


-0-113 


BC 


0-0 


-0-150 


CD 


0-0 


0-0 


DE 


+ 0-057 


-0-051 


EF 


-0-048 


-0-127 


FA 


-0-031 


-0-039 




+ 0-131 


-0-480 




-0-079 


+ 0-052 



*, 1+ K 



/c,/ 2 /s 
+ 0-074 

0-0 
+ 0-093 
+ 0-099 
+ 0-027 
+ 0-037 

+ 0-330 
-0-080 
+ 0-250 



(1) 
(2) 



AA = 8L 



k 2 ld/s 
-0-113 

0-0 

0-0 
-0-088 
+ 0-074 
+ 0-047 

-0-201 
+ 0-121 
-0-080 



81 
-0-039 

0-0 
+ 0-093 
+ 0-011 
+ 0-101 
+ 0-084 

+ 0-289 
-0-039 
+ 0-250 



342 SURVEYING PROBLEMS AND SOLUTIONS 

(4) Bowditch's method 

s d I 8d 81 

AB 514-63 +363-898 +363-898 -0-098 +0-057 

BC 341-36 +341-360 0-0 -0-065 +0-038 

CD 324-15 0-0 -324-150 -0-062 +0-036 

DE 462-37 -231-185 -400-420 -0-088 +0-052 

EF 386-44 -334-667 +193-220 -0-074 +0-043 

FA 217-42 -138-978 +167-202 -0-041 +0-024 

Is 2246-37 -0-428 +0-250 

[id -0-428 A/ +0-250 
Xd 1410-088 2/ 1448-890 

Using the formulae, 

8d =^xs = -°;f* 5 = -1-906 x 10~ 4 5 
2s 2246-37 

81 = *L s = +°' 250s = +1-112 x 10- 4 s 
2s 2246-37 

Ex. Sd, = -1-906 x 10" 4 x 514-63 = -0-098 

81, = +1-112 x 10~ 4 x 514-63 = +0-057 

(5) Transit or Wilson's method 
Using the formulae, 

8d = M. x d = -°' 428<i = -3-035 x 10" 4 d 
2d 1410-088 

5/ = A/ z = +0-250 / = +1 . 725 10 -4 Z 
2/ 1448-890 

Ex. Sd, = -3-035 x 10 -4 x +363-898 = -0-111 

5/, = +1-725 x 10~ 4 x +363-898 = +0-063 





8d 




81 


AB 


-0-111 




+ 0-063 


BC 


-0-103 




0-0 


CD 


-0-0 




+ 0-056 


DE 


-0-070 




+ 0-069 


EF 


-0-102 




+0-033 


FA 


-0-042 




+0-029 




Ad -0-428 


A/ 


+0-250 



TRAVERSE SURVEYS 343 

(6) Smirnoff's method 

N.B. As bearings of BC and CD are 90° and 180° respectively, the 

values of g ^ cos90 > and 5 < sinl80 > will be infinity, 
cos 90 sin 180 

Thus the whole survey is turned clockwise through 20° giving new 

bearings (with an accuracy of ± 10") 









e 


s 










AB 


065-00 514-630 










BC 


110-00 341-360 










CD 


200-00 324-150 










DE 


230-00 462-370 










EF 


320-00 386-440 










FA 


340-16 217-420 






sin0 


8 (sin 
xlO" 


6) 

-6 


^ dS(sin#) 
sin# 


s 


8d 


d (Adj) 


AB 0-906 308 


20 


+ 466-413 +0-010 


0-074 


-0-084 


+ 466-329 


BC 0-939693 


17 


-t 


320-774 +0-006 


0-051 


-0-057 


+320-717 


CD 0-342020 


45 




110-866 -0-013 


0-017 


-0-030 


-110-896 


DE 0-766044 


31 


- 


354-196 -0-014 


0-056 


-0-070 


-354-266 


EF 0-642 788 


37 




248-399 -0-014 


0-039 


-0-053 


-248-452 


FA 0-337643 


46 


+ 


73-410 -0-010 
787-187 
786-871 


0-012 


-0-022 


- 73-432 



Ad + 0-316 0-067 0-249 - 0-316 
2d 1574-058 



cosO 5 ( cos i ) / l8(cos&) z Ss_ Sl /(Adj) 



xlO" 6 cos e 

AB 0-422618 44 +217-492 0-023 0-046 +0-069 +217-561 

BC 0-342020 45 -116-752 0-015 0-025 +0-040 -116-712 

CD 0-939693 17 -304-601 0-006 0-065 +0-071 -304-530 

DE 0-642788 37 -297-206 0-017 0-064 +0-081 -297-125 

EF 0-766 044 31 +296-030 0-012 0-064 +0*076 +296-106 

FA 0-941274 17 +204-652 0-004 0-044 +0-048 +204-700 
+ 718-174 
-718-559 

A/ - 0-385 0-077 +0-385 
21 1436-733 



344 



SURVEYING PROBLEMS AND SOLUTIONS 



Departure 

8s = J_ 



M d8(sin0) ] 
sintf J 

1 [+0-316-0-067] = °' 249 



1574-1 



1574-1 



= 1-581 x 10~ 4 



Latitude 



8s = 1T A/ /a(cosg) 1 
s £/[ cos0 J 



= — L_ [0-385 - 0-077] = -°-^ - 8 - = 2-14 x 10~ 4 
1436-7 1436-7 

The co-ordinates must now be transposed to their original orienta- 



tion. 



Using Eqs. (3.29/30) 





x 2 = 


x, cos 20 


- y, sin 20° 








y 2 = 


y, cos 20 


+ x, sin 20° 








cos 20° = 


0-939 693 










sin 20° = 


0-342020 








Line AB 












x, +466-329 


+ 438-206 


- 74-410 


= +363-796 


8d = 


-0-102 


y, +217-561 


+ 204-441 


+ 159-494 


= +363-935 


81 = 


+0-037 


Line BC 












x, +320-717 


+ 301-376 


+ 39-918 


= +341-294 


8d = 


-0-074 


y, +116-712 


-109-673 


+ 109-692 


= + 0-019 


81 = 


+ 0-019 


Line CD 












x, -110-869 


-104-208 


+ 104-155 


= - 0-053 


8d = 


-0-053 


y, =304-530 


-286-165 


- 37-929 


= -324-094 


5/ = 


+ 0-056 


Line DE 












x, -354-266 


-332-901 


+ 101-623 


= -231-278 


8d = 


-0-093 


y, -297-125 


-297-206 


- 121-166 


= -400-372 


81 = 


-0-048 


Line EF 












x, -248-452 


-233-469 


- 101-274 


= -334-743 


8d = 


-0-076 


y, +296-106 


+ 278-249 


- 84-976 


= +193-273 


81 = 


+0-053 


Line FA 












x, - 73-432 


- 69-004 


- 70-011 


= -139-015 


8d = 


-0-037 


y, +204-700 


+ 192-355 


- 25-115 


= +167-240 


81 = 


+ 0-038 



TRAVERSE SURVEYS 
Analysis of corrections (Figs. 6.27 — 6.31) 



345 



Line AB fs- 514ft J 
045* 00; 




Fig. 6. 27 



thod 


86 






8s 


8d 


5/ 


1 







+0-15 


+0-105 


+0-105 


2 







-0-09 


-0-063 


-0-063 


3a 







-0-07 


-0-047 


-0-047 


3b 







-0-05 


-0-039 


-0-039 


4 


-44" 




-0-03 


-0-098 


+0-057 


5 


-48" 




-0-03 


-0-111 


+ 0-063 


6 


-36" 




-0-05 


-0-102 


+ 0-037 


Line 


BC (S-3AM 
090* 00' 




« 










4 










, 


3o 

o- o— 


— L o 




— _ — 


• > 





1 3b 



6d 



Fig. 6.28 



Method 


86 


8s 


8d 


81 


1 





-0-17 


-0-174 


0-0 


2 





-0-06 


-0-060 


0-0 


3a 





-0-15 


-0-149 


0-0 


3b 





-0-15 


-0-150 


0-0 


4 


+ 22" 


-0-07 


-0-065 


+0-038 


5 





-0-10 


-0-103 


0-0 


6 


+ 11" 


-0-07 


-0-074 


+0-019 



346 



SURVEYING PROBLEMS AND SOLUTIONS 



Line CD (S=-324) 
180* 00' 



6 
4o- 1 



Fig. 6. 29 



Method 


86 


8s 


8d 


81 


1 





+0-17 


0-0 


+ 0-165 


2 





+0-18 


0-0 


+0-181 


3a 





+0-10 


0-0 


+0-096 


3b 





+0-09 


0-0 


+ 0-093 


4 


+ 44" 


0-07 


-0-062 


+0-036 


5 





+0-06 


0-0 


+0-056 


6 


+ 38" 


0-06 


-0-053 


+0-056 



Line Of s = 462 


6/ 




210* 00' 


■0-10 / 


/D 


5 






4 V 




v 


-oio ) 


*\b 


♦0-10 60 


V / 

^r 2 








9C 


' to closing error 



Fig. 6. 30 



Method 


86 


8s 


8d 


81 


1 





0-23 


-0-118 


-0-204 


2 





0-08 


-0-040 


-0-069 


3a 





0-04 


+0-017 


+0-029 


3b 





0-02 


+ 0-006 


+0-011 



TRAVERSE SURVEYS 



347 



(N.B. 90 



thod 86 


8s 


8d 


81 


4 +45" 


0-0 


-0-088 


+0-052 


5 +44" 


0-03 


-0-070 


+0-069 


6 + 26" 


0-09 


-0-093 


-0-048 


° to closing error) 









Line EF s = 386 
300° 00' 
\3o 


61 






3?\? 

\4 




♦01 






5/> 

■ 




F 


i 




-0-1 






♦ 0-1 


6d 


On line of closing 


error 









Method 


86 


8s 


8d 


81 


1 





0-19 


-0-171 


+0-098 


2 





0-12 


-0-187 


+0-108 


3a 





0-23 


-0-202 


+0-117 


3b 





0-19 


- 0-175 


+0-101 


4 





0-09 


- 0-074 


+ 0-043 


5 


-5" 


0-10 


-0-102 


+0-033 


6 





0-09 


-0-076 


+ 0-053 



(N.B. On line of closing error) 

The following may be conjectured: 

(1) The first four methods do not change the bearings. 

(2) Method (1) has a greater effect on the linear values than any other. 

(3) There is no change in bearing in Wilson's method when the line 
coincides with the axes. 

(4) There is little or no change in bearing on any line parallel to the 
closing error in any of the methods analysed - maximum linear correc- 
tion. 

(5) Wilson's method has less effect on the bearings than Bowditch's, 
but more than Smirnoff's. 

(6) The maximum change in bearing occurs at 90° to the closing 
error— maximum linear correction. 



348 SURVEYING PROBLEMS AND SOLUTIONS 

Exercises 6(c) (Traverse Adjustment) 

12. The mean observed internal angles and measured sides of a 
closed traverse ABCDA (in anticlockwise order) are as follows: 

Angle Observed Value Side Measured Length (ft) 

DAB 97° 41' AB 221-1 

ABC 99° 53' BC 583-4 

BCD 72° 23' CD 399-7 

CD A 89° 59' DA 521-0 

Adjust the angles, compute the latitudes and departures assuming 
that D is due N of A, adjust the traverse by the Bowditch method; 
and give the co-ordinates of 3, C and D relative to A. 

Assess the accuracy of these observations and justify your assess- 
ment. (I.C.E. Ans. B -30-3 N, + 219-7 E, 

C +523-9 N, +397-5 E, 
D +522-6 N, - 1-2 E) 

13. The measured lengths and reduced bearings of a closed theodo- 
lite traverse ABCD are as follows: 

Line Length (ft) Bearing 

AB 454-9 Due N 

BC 527-3 Due W 

CD 681-0 S25°18'W 

DA 831-2 N78°54'E 

(a) Adjust the traverse by the Bowditch method and taking A as 
the origin, find the co-ordinates of B, C and D. 

(b) Assess the accuracy of the unadjusted traverse. 

(c) Suggest, and outline briefly, an alternative method of adjusting 
the traverse so that the bearing of AB is unaltered by the adjustment. 

(I.C.E. Ans. B 455-0 N, 0-5 E, 
C 455-1 N, 526-2 W, 
D 160-3 S, 816-5 W) 

14. The following lengths, latitudes and departures refer to a closed 
traverse ABCDEA : 

Length Latitude Departure 

AB 3425-9 3425-9 

BC 938-2 812-6 469-1 

CD 4573-4 2287-1 -3961-0 

DE 2651-3 -2295-7 -1325-9 

EA 1606-4 - 803-0 1391-1 

Adjust the traverse by the Bowditch method, finding the corrected 
latitudes and departures to the nearest 0*1 ft. 



TRAVERSE SURVEYS 349 

Discuss the merits and demerits of this method, with particular 
reference to its effect on lines CD and DE. 

(L.U. Ans. 



AB 


- 0*3, 


+ 3426-1 


BC 


+ 812-5, 


+ 469-2 


CD 


+ 2286-8, 


-3960-7 


DE 


-2295-9, 


-1325-8 


EA 


- 803-1, 


+ 1391-2) 



15. In a closed traverse ABCDEFA the lengths, latitudes and 
departures of lines (in ft) are as follows: 

Line AB BC CD DE EF FA 

Length 1342-0 860-4 916-3 1004-1 1100-0 977-3 

Latitude -1342-00 -135-58 +910-35 +529-11+525-99-483-23 

Departure 0-0 +849-65 +104-26 + 853-38- 966-08-849-42 

Adjust the traverse by the Bowditch method and give the corrected 
co-ordinates of A as (0, 0) 

(L.U. Ans. A 0-0, 0-0 D -569-56, +958-04 

B -1343-00, +1-77 E - 41-20, +1812-74 
C -1479-22, + 852-56 F +483-96, + 848-12 

16. A traverse ACDB is surveyed by theodolite and chain. The 
lengths and quadrantal bearings of the lines, AC, CD and DB are 
given below. 

If the co-ordinates of A are x = 0, y = and those of B are 
x = 0, y = +897-05, adjust the traverse and determine the co-ordinates 
of C and D. 

The co-ordinates of A and B must not be altered. 

Line AC CD DB 

Length 480-6 292-0 448-1 

Bearing N25°19'E N37°53'E N59°00'W 

(L.U. Ans. C +205-2, 435-0 
D +384-4, 665-8) 

17. The lengths, latitudes and departures of the lines of a closed 
traverse are given below. 

In one of the lines it appears that a chainage has been misread by 
40 ft. Select the line in which the error is most likely to have occurred, 
correct it and adjust the latitudes and departures by the Bowditch 
method to the nearest 0-1 ft. 



Line 


Length (ft) 


Latitude 


Departure 


AB 


310-5 


+ 301-2 


+ 75-4 


BC 


695-8 


+ 267-1 


-642-5 


CD 


492-8 


-299-8 


-391-1 



Line 


Length (ft) 


Latitude 


DE 


431-7 


-359-1 


EF 


343-1 


+ 173-5 


FA 


401-9 


- 49-0 



350 SURVEYING PROBLEMS AND SOLUTIONS 

Departure 
+ 239-6 
+ 296-0 
+ 398-9 

(L.U. Ans. Line DE 40 ft too long 
AB +301-1 + 75-6 DE -392-5 +262-0 
BC +267-0 -642-1 EF +173-4 +296-2 
CD -299-9 -390-8 FA - 49-1 +399-1) 

18. (a) Why is the accuracy of angular measurement so important in 
a traverse for which a theodolite and steel tape are used? 

(b) A and D are the terminals of traverse ABCD. Their plane 
rectangular co-ordinates on the survey grid are: 

Eastings Northings 

A +5861-14 ft +3677-90 ft 

D +6444-46 ft +3327-27 ft 

The bearings adjusted for angular misclosure and the lengths of 
the legs are: 

AB 111° 53' 50" 306-57 ft 

BC 170° 56' 30" 256-60 ft 

CD 86° 43' 10" 303-67 ft 

Calculate the adjusted co-ordinates of B and C 

(N.U. Ans. B E 6100*70 N 3563-47 
C E 6141-19 N 3309-99) 

19. From an underground traverse between two shaft-wires, A and D, 
the following partial co-ordinates in feet were obtained: 

AB E 150-632 ft S 327-958 ft 

BC E 528 -314 ft N 82-115 ft 

CD E 26-075 ft N 428 -862 ft 

Transform the above partials to give the total Grid co-ordinates of 
station B given that the Grid co-ordinates of A and D were: 

A E520 163-462 metres N 432182-684 metres 

D E 520 378-827 metres N 432238-359 metres 

(aide memoire : X = *, + k(x-y6) 

Y = y, + k(y + xd)) 
(N.R.C.T. Ans. B E 520 209-364 N 432082-480) 

20. (a) A traverse to control the survey of a long straight street forms 
an approximate rectangle of which the long sides, on the pavements, 
are formed by several legs, each about 300 ft long and the short sides 
are about 40 ft long; heavy traffic prevents the measurement of lines 
obliquely across the road. A theodolite reading to 20" and a tape 



TRAVERSE SURVEYS 351 

graduated to 0*01 ft are used and the co-ordinates of the stations are 
required as accurately as possible. 

Explain how the short legs in the traverse can reduce the accuracy 
of results and suggest a procedure in measurement and calculation 
which will minimize this reduction. 

(b) A traverse TABP was run between the fixed stations T and 
P of which the co-ordinates are: 

E N 

T +6155-04 +9091-73 

P +6349-48 +9385-14 

The co-ordinate differences for the traverse legs and the data from 
which they were calculated are: 

Length Adjusted Bearing AE AN 

TA 354-40 210°41'40" -180-91 -304-75 

AB 275-82 50° 28' 30" +212-75 +175-54 

BP 453-03 20° 59' 50" +162-33 +422-95 

Applying the Bowditch rule, calculate the co-ordinates of A and fi. 

(L.U. Ans. A E 5974-22 N 8786-87 
B E 6187-04 N 8962-33) 

21. The co-ordinates in feet of survey control stations A and B in a 
mine are as follows: 

Station A E 8432-50 N 6981-23 

Station B E 9357-56 N 4145-53 

Undernoted are azimuths and distances of a traverse survey between 
A and B. 

Line Azimuth Horizontal Distance 

4-1 151° 54' 20" 564-31 

1-2 158° 30' 25" 394-82 

2-3 161° 02' 10" 953-65 

3-4 168° 15' 00" 540-03 

4-fi 170° 03' 50" 548-90 

Adjust the traverse on the assumption that the co-ordinates of 
stations A and B are correct and state the corrected co-ordinates of 
the traverse station E N 

(M.Q.B./S Ans. A 8432-50 6981-23 

1 8698-24 6483-55 

2 8842-91 6116-28 

3 9152-83 5214-64 

4 9262-82 4686-07 
fi 9357-56 4145-53) 



352 SURVEYING PROBLEMS AND SOLUTIONS 

22. Define the terms 'error of closure' and 'fractional linear closing 
error' as applied to closed traverse surveys. What error of closure 
would be acceptable for a main road traverse survey underground ? 
Starting with the equations 



/ 


= 


s cos a 


d 


= 


s sin a 


derive the Smirnoff equations 






d JL = lUdL) 

S Zl{ 


V/ d(cosa)) 
cos a j 


and *? = -Lf 


[dD) Sd <*(sina)] 
sin a J 


where a 


= 


bearing angle 


s 


= 


length of traverse draft 


ds/s 


= 


linear precision ratios 


d(cosa) , d(sina) 

ana 

cos a sin a 


= 


angular precision ratios 


11 


= 


sum of latitudes 


Xd 


= 


sum of departures 


dh 


= 


total closing error in latitudes 


dD 


= 


total closing error in departures. 

(N.U.) 



Exercises 6(d) (General) 

23. The following are the notes of a theodolite traverse between the 
faces of two advancing roadways BA and FG, which are to be driven 
until they meet. 

Calculate the distance still to be driven in each roadway. 



Line Azimuth 




Distance (ft) 


AB 267° 55' 




150 


BC 355° 01' 




350 


CD 001° 41' 




315 


DE 000° 53' 




503 


EF 086° 01' 




1060 


FG 203° 55' 




420 

(Ans. BA produced 352-6 ft 
FG produced 916-5 ft) 


24. The following measurements were 


made in a closed traverse, 


ABCD 






A = 70° 45' ; 


D 


= 39° 15' 



TRAVERSE SURVEYS 353 

AB = 400 ft ; CD = 700 ft ; AD = 1019 ft 
Calculate the missing measurements. 

(L.U./E Ans. B = 119° 58', C = 130° 02', 

BC = 351-1 ft) 
25. Particulars of a traverse survey are as follows: 



_ine 


Length (ft) 


Deflection Angle 


AB 


330 


B 76° 23' right 


BC 


515 


C 118° 29' right 


CD 


500 


D 79° 02 'right 


DA 


375 


A 86° 06' right 



Bearing of line AB 97° 15' 

Prepare a traverse sheet and so calculate the length and bearing 
of the closing error. 

(L.U./E Ans. 6*4 ft, N347°04'W) 

26. The interior angles of a closed (clockwise) traverse ABCDEA 
have been measured with a vernier theodolite reading to 20", with re- 
sults as follows: 



Angle at A 


88° 03' 20" 


B 


117° 41' 40" 


C 


126° 13' 00" 


D 


119° 28' 40" 


E 


88° 35' 00" 



Adjust the measurements to closure and find the reduced bearings 
of the other lines if that for line AB is S 42° 57' 20" E. 

(L.U./E. Ans. BC S48°59'40"W DE N14°54'20"W 
CD N 68° 41' 40" W EA N 45° 37' 20" E) 

27. An approximate compass traverse carried out over marshy ground 
yielded the following results: 



Line 


Length (ft) 


Bearing 


AB 


386 


139° 


BC 


436 


50° 


CD 


495 


335° 


DE 


271 


249° 


EA 


355 


200° 



Plot the traverse to a scale of 100 ft to the inch and adjust it graph- 
ically to closure. 

28. A plot of land is up for sale and there is some doubt about its 
area. As a quick check, a compass traverse is run along the boundaries. 
Determine the area enclosed by the traverse from the following data: 



Line 


Bearing 


AB 


195° 


BC 


275° 


CD 


182 1 / 2 ° 


DE 


261^° 


EF 


343° 


FG 


5° 


GH 


8oy 2 ° 


HA 


102 1 / 2 ° 



354 SURVEYING PROBLEMS AND SOLUTIONS 

Feet 

528 

548 

813 
1293 

788 

653 
1421 

778 
(I.C.E. Ans. 56 acres) 

29. The traverse table below refers to a closed traverse run from 
station D, through 0,G and H and closing on D. The whole-circle 
bearing of from D is 06° 26' and G and H lie to the west of the 
line OD. 

Compute the latitudes and departures of 0,G and H with reference 
to D as origin, making any adjustments necessary. 

Observed Internal Angles Length in feet 

DO 547-7 
OG 939-8 
GH 840-2 
HD 426-5 
Ans. +545-1, + 61-0 
G +846-1, -830-7 
H +121-6, -408-2) 

30. The field results for a closed traverse are: 



HDO 


79° 47' 


DOG 


102° 10' 


OGH 


41° 11' 


GHD 


136° 56' 




(I.C.E 



Line 


Whole Circle Bearing 


Length (ft) 


AB 


0°00' 


166 


BC 


63° 49' 


246 


CD 


89° 13' 


220 


DE 


160° 55' 


202 


EF 


264° 02' 


135 


FA 


258° 18' 


399 



The observed values of the included angles check satisfactorily, 
but there is a mistake in the length of a line. 

Which length is wrong and by how much ? 

As the lengths were measured by an accurate 100 ft chain, suggest 
how the mistake was made. 

(I.C.E. Ans. Line BC 20 ft short) 

31. The following traverse was run from station I to station V be- 
tween which there occur certain obstacles. 



TRAVERSE SURVEYS 



355 



Line 


Length (ft) 


Bearing 


I -II 


351-3 


N 82°28'E 


II - HI 


149-3 


N 30°41'E 


III - IV 


447-3 


S 81°43'E 


IV- V 


213-3 


S 86°21'E 



It is required to peg the mid-point of I — V. 

Calculate the length and bearing of a line from station III to the re- 
quired point. (I.C.E. Ans. 171-1 ft S42°28'E) 

32. Two shafts, A and B, have been accurately connected to the 
National Grid of the Ordnance Survey and the co-ordinates of the shaft 
centres, reduced to a local origin, are as follows: 

Shaft A E 10 055-02 metres N 9768-32 metres 

Shaft B E 11801-90 metres N 8549-68 metres 

From shaft A, a connection to an underground survey was made by 
wires and the grid bearing of a base line was established from which 
the underground survey was calculated. Recently, owing to a holing 
through between the collieries, an opportunity arose to make an under- 
ground traverse survey between the shafts A and B. This survey was 
based on the grid bearing as established from A by wires, and the co- 
ordinates of B in relation to A as origin were computed as 

E 5720-8 ft S 4007 -0 ft 

Assuming that the underground survey between A and B is correct, 
state the adjustment required on the underground base line as estab- 
lished from shaft A to conform to the Nation Grid bearing of that line. 

(M.Q.B./S Ans. 00°06'30") 

33. It is proposed to sink a vertical staple shaft to connect X on a 
roadway CD on the top horizon at a colliery with a roadway GH on the 
lower horizon which passes under CD. 

From the surveys on the two horizons, the undernoted data are 
available: 

Upper Horizon 

Station Horizontal Angle 



Inclination 
+ 1 in 200 



Inclined 
Length (ft) 

854-37 



B 



276° 15' 45' 



88° 19' 10' 



+ 1 in 400 943-21 



Remarks 

co-ordinates of A 
E 6549 -10 ft 
N 1356-24 ft 

Bearing AB 
N 30° 14' 00" E 



level 



736-21 



D 



356 



SURVEYING PROBLEMS AND SOLUTIONS 



Lower Horizon 

Station Horizontal Angle Inclination 



F 
G 
H 



193° 46' 45" 
83° 03' 10" 



+ 1 in 50 
+ 1 in 20 
level 



Inclined 
Length (ft) 

326-17 

278-66 

626-10 



Remarks 

co-ordinates of E 
E 7704-08 ft 
N 1210-88 ft 
Bearing EF 
N 54° 59' 10" E 



Calculate the co-ordinates of X 

(M.Q.B./S Ans. X = E 8005-54 ft 

N 1918-79 ft) 

34. Calculate the co-ordinate values of the stations B,C,D and E 
of the traverse ABCDEA, the details of which are given below. 

Data: Co-ordinates of A 1000*0 ft E 1000 -0 ft N 

Bearing of line AB 0°00' 
Length of line AB 342*0 ft 

Interior Angle 
BAE 27° 18' 00" 

CBA 194° 18' 40" 

DCB 146° 16' 00" 
EDC 47° 27' 20" 

AED 124° 40' 00" 

(R.I.C.S. Ans. B 1000-0 E, 1342-0 N C 898-2 E, 1741-1 N 

D 1035-2 E, 2129-6 N E 1313-4 E, 1607-0 N) 

35. The table below gives the forward and back quadrantal bearings 
of a closed compass traverse. 

Tabulate the whole-circle bearings corrected for local attraction, 
indicating clearly your reasons for any corrections. 



Length (ft) 


AB 


342 


BC 


412 


CD 


412 


DE 


592 


EA 


683 



Line 


Length (ft) 


Forward Bearing 


Back Bearing 


AB 


650 


N 55° E 


S54° W 


BC 


328 


S 671/2° E 


N 66° W 


CD 


325 


S 25° W 


N 25° E 


DE 


280 


S 77° W 


N 75 1 / 2 ° E 


EA 


440 


N 64%° W 


s 63y 2 ° e 



A gross mistake of 100 ft has been made in the measurement or 
booking of one of the lines. State which line is in error. Using this 
corrected length, adjust the departure and latitude of each line of the 
traverse to close, using Bowditch's method of adjustment. 



(L.U. Ans. Local attraction at B and E, CD 100 ft too small) 



TRAVERSE SURVEYS 357 

36. It is proposed to extend a straight road AB in the direction AB 
produced. The centre line of the extension passes through a small 
farm and in order to obtain the centre line of the road beyond the farm 
a traverse is run from B to a point C, where A, B and C lie in the 
same straight line. 

The following angles and distance were recorded, the angles being 
measured clockwise from the back to the forward station: 

ABD = 87° 42' BD = 95-2 ft 

BDE = 282° 36' DE = 253-1 ft 

DEC = 291° 06' 
Calculate (a) the length of the line EC 

(b) the angle to be measured at C so that the centre 
line of the road can be extended beyond C. 

(c) the chainage of C taking the chainage of A as zero 
and AB = 362 ft. 

(L.U. Ans. (a) 58-3 ft; (b) 58° 36'; (c) 576-8 ft) 

37. The following are the notes of a traverse made to ascertain the 
position if the point F was in line with BA produced. 

Line Azimuth Distance 

AB 355° 30' 600 ft level 

BC 125° 00' 310 ft rising 1 in 2 

CD 210° 18' 378 ft level 

DE 130° 36' 412 ft level 

EF 214° 00' 465 ft level 

Calculate the difference in the azimuths of AF and BA and the 
extent to which the point F is out of alignment with BA produced. 

(N.R.C.T. Ans. 0°01'; 0-3 ft) 

38. The following notes were made when running a traverse from a 
station A to a station E: 

Side W.C. Bearing Length (ft) 

AB 119° 32' 264-8 

BC 171° 28' 162-4 

CD 223° 36' 188-3 

DE 118° 34' 316-5 
A series of levels were also taken along the same route as follows; 

BS I.S. F.S. R.L. Remarks 

6 * 84 246-20 B.M. near A 

3-86 Sta. A 

11*02 Sta. B 

1*32 13-66 C.P. 1 



358 SURVEYING PROBLEMS AND SOLUTIONS 

BS I.S. F.S. R.L. Remarks 

Sta. C 



I.S. 


F.S. 


9-66 




12-96 






13-44 




12-88 



Sta. D 
0-82 13-44 C.P. 2 

Sta. E 

Calculate the plan length, bearing and average gradient of the line 
AE. (L.U. Ans. 705-1 ft; 145° 11'; 1 in 22-75) 

39. The following are the notes of an underground theodolite traverse. 

Line Azimuth Distance (ft) Vertical Angle 

AB 180° 00' 

BC 119° 01' 181-6 +15° 25' 

CD 160° 35' 312-0 +12° 45' 

DE 207° 38' 320-0 -19° 30' 

EF 333° 26' 200-0 -14° 12' 

It is proposed to drive a cross-measures drift dipping from station 
B at a gradient of 1 in 10 on the line of AB produced to intersect at a 
point X, a level cross-measures drift to be driven from station F. 
Calculate the azimuth and length of the proposed drift FX. 

(Ans. 340° 34'; 83-1 ft) 

40. The following are the notes of an underground theodolite traverse: 
Line Azimuth Distance (ft) Vertical Angle 

350 

190 

600 

155 +28° 

800 -12° 

It is proposed to drive a cross-measures drift to connect stations 
A and F. Calculate the gradient and length of the cross-measures 
drift, and the azimuth of the line FA. 

(M.Q.B./M Ans. 1 in 14-8 (3° 52'); 1391-3 ft (incl); 182°17') 

Bibliography 

MIDDLETON,R.E. and chadwick, O., A Treatise on Surveying (Spon) 

JAMESON, A.H., Advanced Surveying (Pitman) 

CLARK, D., Plane and Geodetic Surveying (Constable) 

RICHARDSON, P., Project Surveying (North Holland Publishing Co.) 

HOLLAND, J. L. WARDELL, K and WEBSTER, A.G., Surveying, Coal 

Mining Series (Virtue) 

SMIRNOFF, M. v., Measurements for Engineering and Other Surveys 

(Prentice-Hall) 

RAINSFORD, HUME F., Survey Adjustments and Least Squares 

(Constable) 



AB 


089° 54' 


BC 


150° 12' 


CD 


180° 00' 


DE 


140° 18' 


EF 


228° 36' 



7 TACHEOMETRY 

The word tacheometry is derived from the Greek tocyus swift, 
^eTpov a measure. This form of surveying is usually confined to the 
optical measurement of distance. 

In all forms of tacheometry there are two alternatives: 

(a) A fixed angle with a variable length observed. 

(b) A variable angle with a fixed length observed. 

In each case the standard instrument is the theodolite, modified to suit 
the conditions. 

The alternatives are classified as: 

(1) Fixed angle: (a) Stadia systems, (b) Optical wedge systems. 

(2) Variable angle: (a) Tangential system - vertical staff, (b) Sub- 
tense system — horizontal staff. 

There are two forms of stadia: 

(1) Fixed stadia, found in all theodolites and levels. 

(2) Variable stadia, used in special tacheometers. 

7.1 Stadia systems — Fixed stadia 

The stadia lines are fine lines cut on glass diaphragms placed 
close to the eyepiece of the telescope, Fig. 7.1. 

Stadia lines 



From Chapter 4, the basic form- 
ulae are: 

D = ms + K (Eq. 4.29) 

= 1 s + (f+d) (Eq. 4.28) 



Fig. 7.1. Diaphragm 




/ 



where m = - = the multiplying constant, 
i 

f = the focal length of the object lens, 

i = the spacing of the stadia lines on the diaphragm, 

d = the distance from the object lens to the vertical axis. 



359 



360 SURVEYING PROBLEMS AND SOLUTIONS 

7.2 Determination of the Tacheometric Constants m and K 

Two methods are available: 

(a) by physical measurement of the instrument itself, 

(b) by reference to linear base lines. 

7.21 By physical measurement of the instrument 

From the general equation, 

D = ms + K 

f 
where m =— and K = f+d. 
i 

In the equation — = — + — (Eq. 4.19) 

f u v 

where u = the distance from the objective to the staff is very large 
compared with / and v and thus 1/u is negligible compared with 
1/v and 1//, 

1 1 

/ v 
i.e. / ^ v 

i.e. fn the length from the objective to the diaphragm with the focus 
at oo. 

With the external focussing telescope, this distance can be chang- 
ed to correspond to the value of u in one of two ways: 

(1) by moving the objective forward, 

(2) by moving the eyepiece backwards. 

In the former case the value of K varies with u, whilst the latter 
gives a constant value. 

The physical value i cannot easily be measured, so that a linear 
value D is required for the substitution of the value of / to give the 
factors i and K, i.e. 

D = sl+U+ d) 
i 

Thus a vertical staff is observed at a distance D, the readings on the 
staff giving the value of s. 

Example 7.1. A vertical staff is observed with a horizontal external 
focussing telescope at a distance of 366 ft 3 in. 
Measurements of the telescope are recorded as: 

Objective to diaphragm 9 in. 

Objective to vertical axis 6 in. 



TACHEOMETRY 

If the readings taken to the staff were 3-52, 5*35 and 7*17 ft, 
calculate 

(a) the distance apart of the stadia lines (i), 

(b) the multiplying constant (m), 

(c) the additive constant (K). 
From Eqs. (4.28) and (4.29), 

D = ms + K 

= l. s + (f+d) 
i 



361 



i = 



fs 



in. 



D-(f + d) 

9-0 x (7-17-3-52) 
366-25 -(0-75+0-50) 

= 9-0 x 3-65 

366-25 - 1-25 
= 0-09 in. 

.-. m = 1 = — = 100 
i 0-09 

K = f+d = 9 in. +6in. = 1-25 ft 

7.22 By field measurement 

The more usual approach is to set out on a level site a base line of 
say 400 ft with pegs at 100 ft intervals. 

The instrument is then set up at one end of the line and stadia 
readings are taken successively on to a staff held vertically at the 
pegs. 

By substitution into the formula for selected pairs of observations, 
the solution of simultaneous equations will give the factors m and K. 

i.e. D, = ms, + K 

D„ = m s„ + K 



Example 7.2 The following readings were taken with a vernier theod- 
olite on to a vertical staff: 



Stadia Readings Vertical Angle 


Horizontal Distance 


2-613 3-359 4-106 0° 


150 ft 


6-146 7-150 8-154 5°00' 


200 ft 


Calculate the tacheometric constants. 





362 



SURVEYING PROBLEMS AND SOLUTIONS 



D = m (4-106 - 2-613) + K = 150 
= l*493m +K =150 

D 2 = m (8-154 - 6-146) cos 2 5°+ Kcos 5° = 200 
= 2-008 mx 0-99620* + 0-996 20 K = 200 

= 1-992 76 m + 0-99620 £ = 200 

Solving these two equations simultaneously gives 
m = 100-05 (say 100) 
K = 0-7 ft 

N.B. The three readings at each staff station should produce a check, 
i.e. Middle-Upper = Lower-Middle 

3-359 - 2-613 = 0-746 4-106 - 3-359 = 0-747 

7-150-6-146 = 1-004 8*154-7-150 = 1-004 

7.3 Inclined Sights 

The staff may be held (a) normal to the line of sight or (b) vertical. 

7.31 Staff normal to the line of sight (Fig. 7.2) 




Fig. 7.2 Inclined sights with staff normal 
As before, D = ms + K 

but H = Dcos0+ SB, 

= Dcos0+ BE sin Q 



i.e. 



H = (ms + K)cos6+ BE sin (9 



(7.1) 
(7.2) 



N.B. BE = h z = staff reading of middle line of diaphragm. BB, is 
- ve when 6 is a depression. 



Vertical difference 



V = Dsin0 



(7.3) 



TACHEOMETRY 



363 



i.e. V = (ms + K) sin0 

As the factor K may be neglected generally, 

H = ms cos 6 + BE sin 
V = ms sin 



(7.4) 

(7.5) 
(7.6) 



If the height of the instrument to the trunnion axis is h^ and the 
middle staff reading ft 2 ,then the difference in elevation 

= h,±V-h z cosd (7.7) 

Setting the staff normal to the line of sight is not easy in practice and 
it is more common to use the vertical staff. 

7.32 Staff vertical (Fig. 7.3) 













S*Ay\ 


A 
















B 




V 










0^ 


^ 


^^ 






V 




^Ke 






10 










HBr 




-<3?r 








H 






/ r 


/>i 












J 














*(ff% 















As before 
i.e. 



but 
thus 



Fig. 7.3 Inclined sights with staff vertical 
D = ms, + K 

= mC4,C,) + K 
A C are the staff readings 
D = m(AC cos0) + K (assuming BA^A = 

BC,C = 90) 



/k 



Also 



Also 



= ms cos + K 


(7.8) 


H = DcosS 




= ms cos z 8 + K cos 


(7.9) 


V = Dsintf 




= ms sin cos + K sin 


(7.10) 


V = H tan (9 


(7.11) 



It can be readily seen that the constant K = simplifies the equations. 
Therefore the equations are generally modified to 



364 SURVEYING PROBLEMS AND SOLUTIONS 

H = ms cos 2 6 (7.12) 

V = V 2 ms sin 26 (7.13) 

If it is felt that the additive factor is required, then the following 
approximations are justified: 

H = (ms +k) cos 2 6 (7.14) 

V = y 2 (ms+K) sin 26 (7.15) 
The difference in elevation now becomes 

= K ±V-h 2 (7.16) 



Example 7.3 A line of third order levelling is run by theodolite, us- 
ing tacheometric methods with a staff held vertically. The usual three 
staff readings, of centre and both stadia hairs, are recorded together 
with the vertical angle (V.A.) A second value of height difference is 
found by altering the telescope elevation and recording the new read- 
ings by the vertical circle and centre hair only. 

The two values of the height differences are then meaned. Com- 
pute the difference in height between the points A and B from the 
following data: 

The stadia constants are multiplying constant = 100. 

additive constant = 0. 

Remarks 
Staff (all measurements in ft) 
Point A 



Point B 



Backsights 


Foresig 


hts 


V.A. Staff 


V.A. 


Staff 


+ 0°02' 6-20 






4-65 






3-10 






+ 0°20' 6-26 








-0°18' 


10-20 
6-60 
3-00 




0°00' 


10-37 



(Aide memoir e: Height difference between the two ends of the theod- 
olite ray = 100s sin0 cos0, where s = stadia intercept and 6 = V.A.) 

(R.I.C.S.) 

V = 100 s sin 6 cos 6 
= 50 s sin 2 6 

To A, V = 50(6-20 - 3*10) sin 0°04' 

= +0-18 ft 



TACHEOMETRY 365 

Difference in level from instrument axis +0*18 

-4-65 





-4-47 


Check reading 




V = 50(3-10) sin0°40' 




= +1-80 




Difference in level from instrument axis 


+ 1-80 




-6-26 




-4-46 


mean 


-4-465 


B, V = 50(10-20 -3-00) sin 


0°36' 


= -3-76 




Difference in level from instrument axis 


- 3-76 




- 6-60 



- 10-36 
Check level -10*37 

mean -10*365 

Difference in level A - B - 10-365 

- 4-465 

- 5-900 ft 



Example 7.4 The readings below were obtained from an instrument 
station B using an anallatic tacheometer having the following con- 
stants: focal length of the object glass 8 in., focal length of the an- 
allatic lens 4*5 in., distance between object glass and anallatic lens 
7 in., spacing of outer cross hairs 0*0655 in. 

Instrument Height of To Bearing Vertical Stadia Readings Remarks 
at Instrument Angle 

B 4*93ft A 69°30' +5° 2-16/3-46/ 4-76 Staff held 

C 159°30' 0° 7-32/9-34/11-36 vertical 

for both 
observa- 
tions 

Boreholes were sunk at A,B and C to expose a plane bed of rock, 
the ground surface being respectively 39-10, 33-68 and 18-45 ft above' 
the rock plane. Given that the reduced level of B was 120*02 ft., de- 
termine the line of steepest rock slope relative to the direction AB. 

(L.U.) 



366 



SURVEYING PROBLEMS AND SOLUTIONS 



By Eq. (4.36), 



Then the multiplying factor m = 



/ = 8in 
/, - 4-5 in 
x = 7 in 
i = 0*0655 in 
//, 



8x4*5 



0-065 5(8 + 4-5-7) 
= 99-93 (say 100) 
At station B: 
To A, H = 100 x 2-60 cos 2 5°= 258*02ft 

V = 258-02 tan 5° = +22-57 ft 

.'. Level of A = 120*02 + 22*57 + 4*93 - 3*46 = 144*06 ft 
To C, H = 100 x 4*04 = 404*00 ft 

V =0 

.'. Level of C = 120*02 + + 4*93 - 9'34 = 115-61 ft 

Dip 



Surface level 120-02 
Depth 33-68 

Bed level 86-34 




Surface level 144-06 
Depth 39-10 

Bed level 104-96 



Surface level 115-61 
Depth 18-45 

Bed level 97-16 



Fig. 7.4 

Gradient AB is (104*96 - 86*34) in 258*02 ft 
18*62 in 258*02 ft 

At point X in Fig. 7.4, i.e. on line AB where the bed level is that of 
C, 

Difference in level AC = 104*96 - 97*16 = 7*80 

258*02 



Length AX = 7*80 x 



18*62 



= 108 *09 ft 



TACHEOMETRY 367 

BX = 258-02 - 108*09 = 149-93 ft 
Angle B = 159° 30' - 69°30' = 90°00' 

In triangle BXC, Angle BCX (a) = tan" 1 BX/BC 

= tan" 1 149-93/404-0 
= 20° 22' 

Therefore the bearing of full dip is perpendicular to the level line CX, 
i.e. 

= Bearing AB + a 
= 69°30'+ 180°+20°22' 
= 269°52' 
7.4 The Effect of Errors in Stadia Tacheometry 

7.41 Staff tilted from the normal (Fig. 7.5) 




Fig. 7.5 Staff tilted from the normal 
If the angle of tilt B is small 

then 4,C, ~ AC = s 

4 t C, = A 2 C 2 cosB 
i.e. s = s, cosjS 

Thus the ratio of error e = — ! — ; — 



= l-cos/8 (7.17) 

Thus the error e is independent of the inclination 0. 

7.42 Error in the angle of elevation with the staff normal 

H = D cos + BE sin (9 

N 



368 



SURVEYING PROBLEMS AND SOLUTIONS 



Differentiating gives — = - D sin d + BE cos Q 
86 

8H = (-D sintf + BE cos 0) Si 



7.43 Staff tilted from the vertical (Fig. 7.6) 



(7.18) 




Fig. 7.6 Staff tilted from the vertical 

Consider the staff readings on the vertical staff at A, B and C, 
Fig. 7.6. If the staff is inclined at an angle /S away from the observ- 
er, the position of the staff normal to the line of collimation will be at 
XY when vertical and X^ when normal to the collimation at the inter- 
section with the inclined staff. 

Assuming that 

BXX, = BYC = B,X,A, = B,Y,Y ^ 90° 
then, with angle 6 an elevation, 

XY = AC cos d = s cos Q 

X,y, = 4,C, cos(6>+/3) = s^cos(d + fi) 

Assuming that XY ~ X, Y, , then 

s cos — s, cos (0 + /S ) 
s, cos (0 + /3) 



s = 



COS0 



(7.19) 



i.e. the reading s on the staff if it had been held vertically compared 
with the actual reading s, taken on to the inclined staff. 



Similarly; if the staff is inclined towards the observer, 

s, cos (0 - /S) 



s = 



COS0 



(7.20) 



TACHEOMETRY 369 

If the angle 6 is a depression the equations have the opposite 
sense, i.e. 

s, cos (0 - B) 
Away from the observer s = (7.21) 

COS0 

^ , , s i cos (.0 + fi) 

Towards the observer s = ' — (7.22) 

cos 8 

Thus the general expression may be written as 

s, cos (0 ±B) 

s = J 1—Ti (7.23) 

cos 6 

The error e in the horizontal length due to reading s, instead of s is 
thus shown as 



True length = H T = ms cos 2 6 

ms, cos (d ± 8) 
cos 6 
Apparent length = H A = ms, cos 2 



cos 2 (7.24) 



Tcos (6 ±B) 1 

Error e = H T -H A = ms. cos 2 (9 ^^ ~ - 1 (7.25) 

L cos 6 J 



Hf — H A 
The error expressed as a ratio = 

ms. cos 2 d 



cos (6 ±8) 



cos 6 



ms^ cos 2 
cos(0 ±B) 



cos 



- 1 (7.26) 



cos 6 cos yS + sin 6 sin /3 - cos 6 
cos 6 
= cos B ± tan sin /3 - 1 
If /3 is small, <5°, then e = B tantf. 

Example 7.5 In a tacheometric survey an intercept of 2*47 ft. was 
recorded on a staff which was believed to be vertical and the vertical 
angle measured on the theodolite was 15°. Actually the staff which 
was 12 ft long was 5 in out of plumb and leaning backwards away from 
the instrument position. 

Assuming it was an anallatic instrument with a multiplying con- 
stant of 100, what would have been the error in the computed horizon- 
tal distance? 



370 



SURVEYING PROBLEMS AND SOLUTIONS 



In what conditions will the effect of not holding the staff vertical 
but at the same time assuming it to be vertical be most serious? What 
alternative procedure can be adopted in such conditions ? 




By Eq. (7.19), 



Thus 



s = 



Fig. 7.7 
s, cos (6 + j8) 



COS0 

/8 = tan -1 5712 x 12 
= 1°59'20" 



s = 



2-47 cos (15° + 1° 59' 20") 



cos 15° 



= 2-4456 



ByEq.(7.12), 



H = ms cos 6 

8H = m cos 2 8s 

= 100 x cos 2 15° x (2-47 - 2'4456) 

= 2-44 cos 2 15 

= 2-28 ft 



Alternatively, 
By Eq.(7.25), 



8H = ms 



a J cos(g + /B ) 

, cos 2 1 



247 cos 2 



-[ 



COS0 

cos 16°59'20' 



cos 15° 
= 230-130 3 [0-99009-1] 
= 2-28 ft 



TACHEOMETRY 371 

7.44 Accuracy of the vertical angle 6 to conform to the overall 
accuracy (Assuming an accuracy of 1/1000) 

From H = ms cos 2 6 

differentiation gives 8H = -2mscos 6 sin 6 86 

3H 1 2ms cos d sin d 86 



For the ratio 



H 1000 ms cos 2 6 

cos 6 



86 = 



2 sin0 x 1000 

1 

cot 6 



2000 



Tf „ „ rtQ 206 265 cot 30° 

If (9=30°, 86 = — seconds 

= 178 seconds; i.e. ~ 3 minutes 

N.B. 1 in 1000 represents 0*1 in 100 ft. The staff is graduated to 
0*01 ft but as the multiplying factor is usually 100 this would repres- 
ent 1ft. 

If estimating to the nearest 0*01 ft the maximum error = ±0*005 ft. 

Thus taking the average error as ±0*0025 for sighting the two 
stadii, 

Average error = 0*0025 V 2 
= ±0*0035 
Error in distance (H) due to reading 

= ±0*0035 m, cos 2 6 
The effect is greater as 6 — * 

Thus, if m = 100, 

8H = ±0*35 ft 

If H = 100 ft 

8H ^ 1 
H 300 

From V = D sin 6 



8 V = D cos 6 86 
8V Dcos6 86 
~V = D sin 6 



= cot0 86 



8V 1 
If 77= 77^^= c ^6 86: 
V 1000 



when = 45°, 



206 265 
86 = ' = 206 sec = 3min 26 sec 



372 SURVEYING PROBLEMS AND SOLUTIONS 

when 6 = 10°, 

3d 



206 265 tan 10° 206 265 x 0-176 3 



1000 



1000 



= 36 sec 



7.45 The effect of the stadia intercept assumption 

(i.e. assuming BA^A = B&C = 90°, Fig. 7.8) 



90-(8+oc) 




90- Oi 



Fig. 7.8 
Let the multiplying factor m = 100 



Then a = tan '- — 



206265 



sec 



200 200 

= 0°17'11'35" 
la = 0°34'23" 



In triangle BA % A A^B = 



In triangle BC^ C BC, = 



s , sin [90 - (6 + a) ] 

sin (90 + a) 
s , cos (d + a) s, (cos d cos a - sin 6 sin a ) 

cos a cos a 

s 2 sin[9O-(0-a)] 

sin (90 -a) 
s 2 cos(0-a) s 2 (cos# cos at + sin sin a) 



cos a 



cos a 



s, (cos 6 cos a - sin 6 sin a ) s,(cos 6 cos a + sin sin a ) 
1 T cos a cos a 

= s (cos - sin tan a) + s 2 (cos + sin 6 tan a ) 

i4 1 C 1 = (s, + s 2 ) cos + (s 2 - s,)(sin tana) (7.27) 



TACHEOMETRY 373 

Thus the accuracy of assuming A X C^ = AC cos 6 depends on the sec- 
ond term (s 2 - s,) (sin 6 tana). 

Example 7.6 (see Fig. 7.8) 

If 6 = 30°, FB = 1000 ft., m = 100 and K = 0, 

^ B = BC < = loo = 5 ' 0ft 



s. = 





i4,B 




cos - sin tan a 
5-0 




0-866 - 0-5 x 0*005 
5-0 





0-866 - 0-0025 
5-790 4 



Similarly 



BC, 



cos 6 + sin tan a 
5-0 



0-866 + 0-0025 

= 5-757 1 
Therefore the effect of ignoring the second term 

(s 2 -s;) (sin tana) = (5-790 4 - 5-757 1) (0-002 5) 
- -0-033 3x0-0025 
= -8-325 x 10~ 5 
The inaccuracy in the measurement FB thus 

= -8-325 x 10~ 2 
~ 0-1 ft in 1000 ft 

and the effect is negligible. 

Thus the relative accuracy is very dependent on the ability to es- 
timate the stadia readings. For very short distances the staff must be 
read to 0*001 ft, whilst as the distances increase beyond clear read- 
ing distance the accuracy will again diminish. 

Example 7.7 

A theodolite has a tacheometric multiplying constant of 100 and 
an additive constant of zero. The centre reading on a vertical staff 
held at a point B was 7*64 ft when sighted from A. If the vertical 



374 



SURVEYING PROBLEMS AND SOLUTIONS 



angle was +25° and the horizontal distance A B 634*42 ft, calculate 
the other staff readings and show that the two intercept intervals are 
not equal. 

Using these values, calculate the level of B if A is 126*50 ft 
A.O.D. and the height of the instrument 4*50 ft. 

(L.U.) 



90-(B + oe) 



90+<* 




90-otT 



Horizontal distance 



Inclined distance 



mcos 2 d 100 cos 2 25° 
= 7-72 ft 

= HD sec 6 

= 634-42 sec 25° 

= 700-00 ft 

= 3-50 

s n cosa 



cos (0 + a) 
3-50 cos 0°17'11" 
cos 25° 17 '11" 
= 3-87 



(sine rule) 



TACHEOMETRY 



375 



Similarly, 


S 2 


s cos a 


cos(0-a) 






3-50 cos 0°17'11" 






cos 24°42'49" 






= 3^85 


Check 3-87 + 3-85 




= 7-72 ft 


Staff readings are 




7-64 7-64 
+ 3-87 and -3-85 


Upper 




11-51 Lower 3-79 
-3-79 


Check s 




7-72 


Vertical difference 




= HDtand 

= 634-42 tan 25° 

= +295-84 ft 


Level of 


A 


= 126-50 
+ 295-84 
+ 4-50 




+ 426-84 






- 7-64 


Level of 


B 


= +419-20 



(sine rule) 



Example 7.8 Two sets of tacheometric readings were taken from an 
instrument station A, the reduced level of which was 15-05 ft., to a 
staff station B. 

(a) Instrument P — multiplying constant 100, additive constant 
14-4 in, staff held vertical. 

(b) Instrument Q — multiplying constant 95, additive constant 
15-0 in, staff held normal to line of sight. 

Inst At To Height of Inst. Vertical Angle Stadia Readings 
P A B 4-52 30° 2-37/3-31/4-27 

Q A B 4-47 30° 

What should be the stadia readings with instrument Q? (L.U.) 

To find level of B (using instrument P) 
By Eq. (7.10), 

V = ms sin B cos 6 + K sin 6 



376 SURVEYING PROBLEMS AND SOLUTIONS 

V, = 100 x (4-27 - 2-37) sin 30° cos 30° +1-2 sin 30 
= 190 x 0-5x0-8660+ 1-2x0-5 
= 82-27 + 0-60 = 82-87 ft 

By Eq. (7.9), tf, = ms cos 2 + K cos 

= 190 x 0-866 03 2 + 1-2 x 0-866 03: 
= 142-49+ 1-03 - 143-52 ft 

Also by Eq. (7.11), V,= // 1 tan0 

= 143-52x0-57735 = 82-86 ft (Check) 

Level of B = 15-05 + Ht of inst + V - middle staff reading 

= 15-05 +4-52+ 82-87 -3-31 = 99-13 ft 

Using instrument Q 
In Fig 7. 2, 

V = (H - BE sin 6) tan 6 

V 2 = (143-52 - BE sin 30°) tan 30° 

= 143-52 x 0-577 35 - BE x 0-5 x 0-577 35 
= 82-86 - 0-28868 BE 
Level of B = 15-05 + 4-47 + V 2 - BE cos = 99-13 

= (82-86 -0-288 68 BE) -0-866 03 BE = 79-61 

-1-15471 BE =-3-25 

BE = -3-25 

1-154 71 

middle reading =■■ 2-81 

By Eq. (7.5), H 2 = ms cos 6 + BE sin d 

= 95 x 0-866 03 s+ 2-81x0-5 = 143-52 
= 82-27 s = 143-52- 1-40 
142-12 



s = 



= 1-727 



82-27 
l / 2 s = 0-86 
'.Readings are 2-81 ±0-86 = 1-95/2-81/3-67 



Example 7.9 Three points A,B and C lie on the centre line of an 
existing mine roadway. A theodolite is set up at 6 and the following 
observations were taken on to a vertical staff. 



TACHEOMETRY 



377 



Staff at Horizontal Vertical Staff Readings 

Circle Circle Stadia Collimation 

A 002°10'20" +2°10' 6-83/4-43 5-63 

C 135°24'40" -1°24' 7*46/4-12 5-79 

If the multiplying constant is 100 and the additive constant zero 
calculate: 

(a) the radius of the circular curve which will pass through A,B 
and C. 

(b) the gradient of the track laid from A to C if the instrument 
height is 5-16. 

(R.I.C.S.) 




Fig. 7.10 



Assumed bearing BA = 002°10'20" 

BC = 135°24'40" 

Angle ABC = 133° 14' 20" 

Angle AOC = 360 - 2(133° 14 '20") 
= 93°31 , 20" 
Line AB 

Horizontal length (H) = ms cos Q 

= 100(6-83-4-43) cos 2 2° 10' 
= 240 cos 2 2° 10' 
- 239-66 ft 



378 SURVEYING PROBLEMS AND SOLUTIONS 

Vertical difference (V) = H tand 

= 239-66 tan 2° 10' 
= +9-07 ft 



Line BC 



H = 100(7-46 -4-12) cos 2 1°24' 

= 333-80 ft 

V = 333-80 tan 1° 24' 

= 8-16 ft 



In triangle ABC 



A-C a- c A + C 
Tan — -— = tan 



a + c 2 

333-80 - 239-66 180 - 133° 14 '20" 



333-80 + 239-66 
94-14 



tan 



573-46 
A - C 

= 4°03'35" 

2 

A -i- C 

^—^ = 23°22'50" 
2 

A = 27° 26 '25" 

C = 19°19'15" 

AB 



tan 23°22'50' 



= 2R (sine rule) 



sin C 

239-66 



R = 



Differences in level 



2 sin 19° 19 '15" 
= 362-18 ft 

BA = 5-16 + 9-07-5-63 
= +8-60 

BC = 5-16-8-16-5-79 

= -8-79 
AC = 17-39 

Length of arc AC = 362-18 x 93°31' 20" rad 

= 362-18 x 1-632 271 
= 591-18 ft 



TACHEOMETRY 



379 



Gradient = 17-39 ft in 591-18 ft 
= 1 in 34 



Example 7.10 The following observations were taken during a tacheo- 
metric survey using the stadia lines of a theodolite (multiplying const- 
ant 100, no additive constant.) 

Station Set Station Staff Readings Vertical Bearing 

at Observed U M L Angle 

B A 5-62 6-92 8-22 +5°32' 026°36' 

C 3-14 4-45 5-76 -6°46' 174°18' 

Calculate (a) the horizontal lengths AB and BC. 

(b) the difference in level between A and C. 

(c) the horizontal length AC. 
Line BA 





s 


= 


8-22-5-62 = 2-60 






Horizontal length 


= 


100 s cos 2 6 




i 


1 */ \ 






= 


100 x 2-60 cos 2 5< 
257-58 ft 


D 32' 




L $/ i 

y 1 


Vertical difference 


= 


HtanO 

257-58 tan 5°32' 

+ 24-95 ft 






if \ 


Line BC 
















s 


= 


5-76-3-14 = 2-62 


8< 


( Jl47 # 42' / 




H 


= 


100x2-62 cos 2 6* 
258-36 ft 


>46' 




\ «■* i 




V 


= 


258-36 tan 6°46' 
-30-66 ft 






\ -* > 

\* 1 

\ ^* ' 
\ jo | 


Native levels 


A 
A 




+ 24-95 
- 6-92 
+ 18-03 above 


B 




\<p / 
\& i 




-30-66 










- 4-45 






c 




C 




-35-11 below 


B 




Fig. 7.11 








Difference i 


n level A-C 53' 14 ft 



380 SURVEYING PROBLEMS AND SOLUTIONS 

In triangle ABC, 

A-C 258-36 - 257-58 180 - 147°42 / 

Tan 2 = 258-36 + 257-58 ta " 2 

A-C 



2 
A + C 



0°01' 



16°09' 



A = 16° 10' 
Length AC = 258-36 sinl47°42' cosec 16°10' = 4 95-82 

Exercises 7(a) 

1. P and Q are two points on opposite banks of a river about 100 yd 
wide. A level with an anallatic telescope and a constant of 100 is set 
up at A on the line QP produced, then at B on the line PQ produced 
and the following readings taken on to a graduated staff held vertically 
at P and Q. 

What is the true difference in level between P and Q and what is 
the collimation error of the level expressed in seconds of arc, there 
being 206 265 seconds in a radian. 





To 


Staff readings in feet 


From 


Upper Stadia 


Collimation 


Lower Stadia 


A 


P 

Q 


5-14 
3-27 


4-67 
1-21 


4-20 
below ground 


B 


P 

Q 


10-63 
5-26 


8-51 
4-73 


6-39 
4-20 



(I.C.E. Ans. 3-62 ft; 104" above horizontal) 

2. Readings taken with a tacheometer that has a multiplying const- 
ant of 100 and an additive constant of 2*0 ft were recorded as follows: 



Instrument at 



Staff at 



Vertical Angle 



Stadia Readings 



Remarks 



Q 



30°00' elevation 



5-73, 6-65, 7-57 



Vertical 
staff 



Although the calculations were made on the assumption that the 
staff was vertical, it was in fact made at right angles to the collimation. 

Compute the errors, caused by the mistake, in the calculation of 
horizontal and vertical distances from the instrument to the foot of the 
staff. Give the sign of each error. 

If the collimation is not horizontal, is it preferable to have the 
staff vertical or at right angles to the collimation? Give reasons for 



TACHEOMETRY 



381 



your preference. 

(I.C.E. Ans. Horizontal error -24*7 ft, Vertical error -13'2 ft) 
3. The following readings were taken with an anallatic tacheometer 
set up at each station in turn and a staff held vertically on the forward 
station, the forward station from D being A. 



Station 


Height of 
Instrument 


Stadia Readings 


Inclination (elevation 


+ ve) 


A 
B 
C 
D 


4-43 
4-61 
4-74 
4-59 


4-93 3-54 2-15 
5-96 4-75 3-54 
5-15 3-72 2-29 
6-07 4-64 3-21 


+ 0°54' 
-2°54' 
+ 2°48' 
-1°48' 



The reduced level of A is 172-0 ft and the constant of the tacheo- 
meter is 100. 

Determine the reduced levels of B, C and D, adjusted to close 
on A, indicating and justifying your method of adjustment. 

(I.C.E. Ans. 177-5; 165-4; 180-7) 

4. The focal lengths of the object glass and anallatic lens are 5 in 
and 4y 2 in respectively. The stadia interval was 0-lin. 

A field test with vertical staffing yielded the following: 
Instrument Staff Staff Vertical Measured Horizontal 

Station Station Intercept Angle Distance (ft) 

P Q 2-30 +7°24' 224-7 

R 6-11 -4°42' 602-3 

Find the distance between the object glass and anallatic lens. 
How far and in what direction must the latter be moved so that the 
multiplying constant of the instrument is to be 100 exactly. 

(L.U. 0*02 in away from objective) 

5. Sighted horizontally a tacheometer reads r, = 6'71 and r 3 = 8'71 

on a vertical staff 361*25 ft away. The focal length of the object glass 

is 9 in. and the distance from the object glass to the trunnion axis 6 in. 

Calculate the stadia interval. (I.C.E. Ans. 0*05 in) 



6. With a tacheometer stationed at X sights were taken on three 
points, A, B, and C as follows: 



nstrument 


To 


Vertical 


Stadia 


Remarks 


at 




Angle 


Readings 




X 


A 


-4°30' 


7-93/6-94/5-95 


R.L. of A = 357-09 (Staff nor- 
mal to line of sight 




B 


0°00' 


4-55/3-54/2-54 


R.L. of B= 375-95 (Staff 

vertical) 




C 


+ 2°30' 


8-85/5-62/2-39 


Staff vertical 



382 



SURVEYING PROBLEMS AND SOLUTIONS 



The telescope was of the draw-tube type and the focal length of 
the object glass was 10 in. For the sights to A and J5, which were of 
equal length, the distance of the object glass from the vertical axis 
was 4*65 in. 

Derive any formulae you use. Calculate (a) the spacing of the 
cross hairs in the diaphragm and (b) the reduced level of C. 

(L.U. Ans. 0-102 in.; 401-7 ft) 

7. The following readings were taken on a vertical staff with a tach- 
eometer fitted with an anallatic lens and having a constant of 100: 

Staff Station Bearing Stadia Readings Vertical Angle 
A 27°30' 2-82 4*50 6*18 + 8°00' 

B 207°30' 2-54 6*00 9-46 -5°00' 

Calculate the reduced levels of the ground at A and B, and the 
mean slope between A and B. 

(L.U. Ans. +41-81; -66-08; 1 in 9*42) 



8. Tacheometric readings were taken from a survey station S to a 
staff held vertically at two pegs A and B, and the following readings 
were recorded: 

Point Horizontal Circle Vertical Circle Stadia Readings 



62°00' 
152°00' 



+ 4° 10 '30" 
-5°05'00" 



4-10/6-17/8-24 
2-89/6-17/9-45 



The multiplying constant of the instrument was 100 and the addi- 
tive constant zero. Calculate the horizontal distance from A to 8 
and the height of peg A above the axis level of the instrument. 

(I.CE. Ans. 770-1 ft; 23-89 ft) 



9. In a tacheometric survey made with an instrument whose constants 
were f/i :*= 100, (f+d) = 1*5, the staff was held inclined so as to be 
normal to the line of sight for each reading. How is the correct inclin- 
ation assured in the field? 

Two sets of readings were as given below. Calculate the gradient 
between the staff stations C and D and the reduced level of each. 
The reduced level of station A was 125*40 ft. 



Instrument 


Staff 


Height of 


Azimuth 


Vertical 


Stadia Readings 


at 


at 


Instrument 




Angle 




A 


C 


4-80 


44° 


+ 4°30' 


3-00/4-25/5-50 




D 




97° 


-4°00' 
(L. 


3-00/4-97/6-94 
U. Ans. 1 in 6-57) 



TACHEOMETRY 



383 



10. (a) A telescope with tacheometric constants m and c is set up 
at A and sighted on a staff held vertically at B. Assuming the usual 
relationship D = ms + c derive expressions for the horizontal and 
vertical distances between A and B. 

(b) An instrument at A, sighted on to a vertical staff held at B 
and C, in turn gave the following readings: 



Sight 



Horizontal Circle 



05°20' 
95°20' 



Vertical Circle 



+ 4°29'00" 
-0°11'40" 



Staff Readings (ft) 



1-45/2-44/3-43 
2- 15/3- 15/4- 15 



If the instrument constants are m = 100, c = 0, calculate the gradient 
of the straight line BC. 

(N.U. Ans. 1 in 16-63) 

7.5 Subtense systems 

7.51 Tangential method (with fixed intercept s and variable vertical 
angles a and /S) 




From Fig. 7.12 



Fig. 7.12 Tangential method 

DC = y = //tana 
AD = s + y = H tan /8 
AC = s = H (tan - tana) 
s 



H = 



Alternatively, as 



tan /3 - tan a 
7 = /3-a, 
s ///cos a 



(7.28) 



sin 7 sin(90-j8) 

H = s cos a cos /8 cosec y 



(7.29) 



384 



SURVEYING PROBLEMS AND SOLUTIONS 



This equation (7.29) was modified by M. Geisler (Survey Review, Oct. 
1964) as follows: 

s 

H = cos a cos (y + a) 

siny 

s cos (2 a + y) + cos y 

siny 



siny 



1 + cos y) - { 1 - cos (2 a + y) } 



ay • 2 

cos -i- - sin 
2 



y y 

2 sin— cos— L 
2 2 



= -s cot— - —s cot- ' 
2 2 2 2 



y 

As y is small, cos 2 -^ ^ 1. 




y 

Also a+-y = 



i y l y 2 

H = -~s cot« - ys cot-j sin 

i y 

= -^scot^- (l-sin 2 0) 

1 y 

= — s cot— cos 2 

2 2 



(7.30) 



Alternatively, the above equation may be derived by reference to 
Fig. 7.13. 

1 y 

H x= ^ s i cot ~2~ where s i = ^i C i 

s, ~ s cos 6 (assuming A,AB and BC X C are 
similar figures) 

H = H.cosd 

1 y 

H = —s cot— cos 2 6 as above 



but 
and 



1 y 

N.B. In the term -^s cot-~ 



.2/ 7^ 
sin (a + ~2 J 



2 y 

COS 4r 



cot -^ is very large, 



TACHEOMETRY 



385 




Fig. 7.13 
so that any approximation to cos 2 Z is greatly magnified and the 

following approximation is preferred: 

As before H = —r—r, cos a cos ( y + a) 



siny 
s 



■cos 



K)~K) (—-?) 



but cosy 



~ 1 



sin y \ II \ 2 

s f cos 20 + cosy 
siny [ 

r 2 



i 



~i 



H 



H 



[" cos 20+1 1 



siny 

s cos 2 ^ 
siny 
s cosecycos 2 



(7.31) 

Geisler suggests that by using special targets on the staff, thus 
ensuring the accuracy of the value of s, and the use of a 1" theodol- 
ite, a relative accuracy up to 1/5000 may be attained. He improved the 
efficiency of the operation by using prepared tables and graphs rela- 
tive to his equation. 

The accuracy of the method is affected by: 

(1) An error in the length of the intercept s. 

(2) An error in the vertical angle. 

(3) Tilt of the staff from the vertical. 



386 SURVEYING PROBLEMS AND SOLUTIONS 



(1) Error in the intercept s 

This depends on (a) error in the graduation, (b) the degree of 
precision of the target attachment. 

8H = ^-^ (7.32) 

s 

(2) Error in the vertical angle 
From Eq. (7.28), H 



^ 



tan jS - tan a 
+ s sec a 8a 



(tan /3 - tan a ) 



-s sec 2 /8 8B 

8H„ = , — ^— £- (7.33) 

P (tanjS-tana) 2 



total r.m.s. error 8H 

(tan/3 - tana) 



= y/8H* + S/J 2 = Jse^a&^ + sec^S/S 2 ] 1 

" /'fan R _ tan^ 2 



If 5a = 83, 



s 5a 

^ = -r. — a — : r; [sec 4 a + sec 4 B ] 

(tan B - tan a) 2 ^ 



-^f.[sec 4 a + sec* BY (7.34) 



If a and B are small, 



yj2H z 8a 
8H = V s (7.35) 

(3) Tilt of the staff 

The effect here is the same as that described in Section 7.43, i.e. 

s,cos(0 ±B) 

s = 

cos 

where B = tilt of the staff from the vertical. 

Example 7.11 A theodolite was set over station A, with a reduced 
level of 148-73 ft A.O.D., the instrument height being 4-74ft. Obser- 
vations were taken to the 10 ft and 2 ft marks on a staff held vertical 
at three stations with the following results: 

Instrument Station Station Observed Vertical Angles 

Top Bottom 
A B +9°10' +3°30' 

A C +1°54' -2°24' 

A D -5°15' -12°10' 



TACHEOMETRY 387 

Find the distance from A to each station and also their reduced 
levels. 

(E.M.E.U.) 

By Eq. (7.28), 

Horizontal distance AB = 



tan /8 - tan a 
8 



Vertical distance 
Level of B 



Horizontal distance AC = 



Vertical distance 
Level of C 





tan9°10' -tan3°30' 




8 




0-100 21 


= 


Afitan3°30' = 4-883 ft 


= 


4-88 + 4-74+ 148-73 -2-00 ft 


= 


156-35 ft A.O.D. 




8 




tanl°54' +tan2°24' 




- 106-5=1 ft 




0-07509 


= 


-106-55 tan 2° 24' = -4-47 ft 


= 


148-73 + 4-74 - 4-47 - 2-00 



147-00 ft A.O.D. 



8 
Horizontal distance AD = 



tanl2°10'-tan5 o 15' 
8 



0-12371 



= 64-53 ft 



Vertical distance = -64-53 tan 12°10' = -13-91 ft 

Level of D = 148-73 + 4-74 - 13-91 - 2-00 ft 

= 137-56 ft A.O.D. 

Alternative solutions 

ByEq.(7.29), 

H 8cos9°10' cos3°30' 

Horizontal distance AD = — : — ,„.,-, ^^^ 

sin (9° 10 - 3°30 ) 

8 x 0-987 23 x 0-998 14 

0-098 74 

= 79-84 ft 

or by Eq. (7.30), AB = % x 8 x cot 2°50' x cos 2 V 2 (12°40' ) 

= 4cot2°50' cos 2 6 o 20' 



388 



SURVEYING PROBLEMS AND SOLUTIONS 

= 4x20-2056x0-99390 
= 79-84 ft 



7.52 Horizontal subtense bar system (Fig. 7.14) 




Theodolite 



Horizontal angle (oc) measured by the theodolite 




Fig. 7.14 Horizontal subtense bar system 

The horizontal bar of known length b, usually 2 metres, is set 
perpendicular to the line of sight TB. Targets at A and C are 
successively sighted and the angle a, which is measured in the 
horizontal plane, recorded. 

The horizontal distance TB, = H is then obtained 



b a 

H = - cot - 

2 2 



If the bar is 2 metres long, 



H = cot— metres 



(7.36) 



(7.37) 



The horizontal angle a is not dependent on the altitude of the bar 
relative to the theodolite. 

N.B. As the bar is horizontal, readings on one face only are necess- 
ary. 

Factors affecting the accuracy of the result are 
(1) The effect of an error in the subtended angle a 

By Eq.(7.36), 

b a 
H = j cot - 



TACHEOMETRY 389 

b 



2 tan — 



a a 
If a is small, then tan——— radians. 



a 
Differentiating with respect to a, 



b 
H = - (7.38) 



SOa - ~ (7-39) 



ft 
but a=- 



8Ha = -uya (7 40) 



b 

lhe error ratio -rr- = 

n a 

where 8a and a are expressed in the same units. 

If 8a = +1" and b = 2m, 

then by Eq. (7.40), 

8// = ± metres 

2 x 206 265 



(7.41) 



8H = ± 412^0 metreS (742) 

* Slf^oo metres (743) 



Example 7.12 To what accuracy should the subtense angle a be 
measured to a bar 2 metres long if the length of sight is approximately 
50 metres and a fractional error of 1/10 000 must not be exceeded? 

By Eq. (7.41), 

SH 8a 1 



but 



H a 10000 

a 
5a = 10000 

b 
a =ll 



8a 10000// 



390 



SURVEYING PROBLEMS AND SOLUTIONS 



2 x 206 265 
= ^OOOOxSO 8600 ^ 5 

= + 0-825 " 

Example 7. 13 If the measured angle a is approximately 2°, how 
accurately must it be recorded if the fractional linear error must not 
exceed 1/10 000? 



8H 


8a 




1 


H 


a 




10000 




8a 


= 


2 x 3600 
10 000 






= 


±0-72" 


an 


error in 


th 


e length o 




H 


= 


b 
a 




8H b 


= 


8b 
a 




H 


= 


8b 
b 



(2) The effect 
From Eq.(7.38), 



(7.44) 

(7.45) 
If the error ratio is not to exceed 1/10000, then 

,?3 = §k 1 

H b 10000 

As the bar = 2 metres = 2000 mm, 8b is limited to 0-2 mm. 

The bar is usually made of invar steel and guaranteed by the manu- 
facturer to ±0-05 mm. It requires a change of 20°C for these limits 
and thus for most uses the bar is considered constant. 

(3) The effect of an error in the orientation of the bar 
Consider half the bar, Fig. 7. 15. 




Fig. 7. 15 Orientation of the bar 



TACHEOMETRY 391 

Let the half bar AB be rotated through an angle 6 to A : B. The line 
of sight will thus be assumed to be at A z . 

r, b . e u . 

AAi = 2x — sin- = osm— 



AA 2 = 8H 



AA y sin 



M 



sin T 



osin-^jsin ycos^- 



6 . j8 
+ cos«- sin-|j- 



sin -£- 
2 

= b sin z x-cot— + b sin-^ cos ■ 
2 2 2 2 



P 



but bcot±-~2f/ 

2 



Q J. 

8H = 2H sin 2 - +- sin0 (7.46) 



Neglecting the second term as both —and 6 are small, 



-f =2sin 2 -^ (7.47) 

If the relative accuracy is limited to 1/10000, then 

m 1 2sin 2 -2- 

H 10000 2 

sin 2 = 5 x 10~ 4 

$ = 1°17' 

As the bar usually has a sighting device, it can be oriented far 
more accurately than to the above limit, and this non-rigorous analysis 
shows that this effect can be ignored. 

The accuracy of the whole system is thus entirely dependent on 
the angle a . 

Assuming the angle a can be measured to ±1" and the bar is 2 
metres, 

da 1 



a 10000 

a = 10000' 



392 



SURVEYING PROBLEMS AND SOLUTIONS 



H = 



2 x 206 265 



a " 10000 
= 41-25m 



metres 



For most practical purposes, for an accuracy of ±lcm, the dis- 
tance can be increased to 75 metres. 

To increase the range of the instrument various processes may be 
used, and these are described in the next section. 

7.6 Methods used in the.field 
7.61 Serial measurement (Fig. 7.16) 




Fig. 7.16 Serial measurement 

tj 2 = //, +h 2 + ... = h 

b 
= ~2l 



By Eq. (7.40), 



8H, = 



Total r.m.s. error = y/2*(5H) z = 



[ cot T +cot f + -] 

H, 8a i 



H;8a 2 



+... 



H 



If ff, =H 2 = H n = — , a, = a 2 = a n and 8a, = 8a 2 = 8 n , 



2SH = ±\/n 



H 2 8a 



n z b 



= +■ 



H 2 8a 

n^b 



If b = 2 metres, 5a = ±1" and f/ = n//, 
Total 8H = ± 



2 x 206 265 x n 3 



= ± 



412 530 n 3 



(7.48) 



(7.49) 



(7.50) 



The error ratio 



TACHEOMETRY 
H Z 



+ 



400000n 3/2 
8H „ H 
H 400000/x 372 



8H 



If 5a = ±1", b = 2 metres, n = 2 and ~ = 1/10000 



then 



/7 



400000 2 10000 



400000x2-83 

H = = 113 metres 

10000 



393 

(7.51) 
(7.52) 



7.62 Auxiliary base measurement (Fig. 7.17) 

For lines in excess of 150m, an auxiliary base of 20 — 30 m may 
be set out at right angles to the traverse line, Fig. 7.17. 




Fig. 7.17 Auxiliary base measurement 
Angles a and /3 are measured 



Differentiating, 



% 


b a 

= — cot — 

2 2 


H 


= }% cotjS 




b a n 
= ~2 cot 2 Cot ^ 



= — cotjS 



8H n 



Sty = 



bcotjS 



5a 



Z?cosec 2 j3 5/3 



(7.53) 
(7.54) 



394 



SURVEYING PROBLEMS AND SOLUTIONS 



Total r.m.s. error 8H 
If a and /8 are both small, 
8H = 



b 2 cot 2 £ 5a bcosec 4 ft 8(3 2 



but 



H = 



8H = 



b 8a 




b 2 8fi z * 


I a A (3 2 


+ 


a 2 F J 


b 






a/3 






~H 2 8a 2 

2 

a 


+ 


H 2 8fi 2 
P 2 J 



(7.55) 



(7.56) 



(7.57) 



If a = /8 and 5a = 5j3, 

8H = 

b H 

As H = - and H = -2- 

* a 

and as a = j8, 



\/2 # Sa 



and 



^ = A 

H, = s/(bH) 
b 



8H = 



y/(bH) 
V2f/ 3/2 Sa 



V H 



(7.58) 



(7.59) 



(7.60) 



If b = 2 metres and 8a = ±1", 



8H = 



H 3 



and the fractional error 



If __ = 1/10 000, 
H 



206 265 
H " 206 265 



(7.61) 
(7.62) 



then 



TACHEOMETRY 

y/H = 20-626 5 
H = 410 metres 



395 



the sub-base H b = \/(2H) 

= V(2x410) 
= 28*7 metres 



7.63 Central auxiliary base (Fig. 7.18) 

For lines in excess of 400 metres, a double bay system may be 
adopted with the auxiliary base in the middle. 




Fig. 7.18 Central auxiliary base 
Length TJ 2 = H = H, + H z 

b ,ol , a b ^ a. , _ 
= — cot — cot p. + — cot — cot /3„ 

2 2 ! 2 2 2 

= ^cot^[cota + cotaj 

2 2' 2 

= *[cot# + cot/6 2 ] 
a 

If a , j8, and j8 2 are each small, 
Differentiating, 



(7.63) 
(7.64) 

(7.65) 



"*».--£#. 



396 



SURVEYING PROBLEMS AND SOLUTIONS 
b 



8Hn = --^r<5)3 2 

"2 



a/3 2 2 



Total t.m.s. error 8H = y/{8fg +8H/3* + 8Hfi 






If a = ft = j^ and 5a = Sft = <5/8 2 , then 

b /fea 2 /2 2 1 1 
as\ (a \a a a 



V6 b8a 



but by Eq. (7.65) 



H = 



a = 



8H = 



2b 

a 2 

lb 



(ff, = #2) 



^/6H 3/2 b8a 

2f /2 ff 2 

V3 fl 3/2 Sa 

2yjb 



If b= 2m and 5a = ±1", 



V3/T 



2V2x 206 265 



H 



3/2 



336 818 

,3/2 



If SH/ff = 1/10000, 



<5tf z. 



8H 



W 



350 000 



y/H 



10000 350000 

V^ - 35 
H ~ 1225 metres 

The auxiliary base f^ = H b = \/H 

= 35 metres. 



(7.66) 



(7.67) 



(7.68) 



(7.69) 
(7.70) 



TACHEOMETRY 



397 



7.64 Auxiliary base perpendicularly bisected by the traverse line 

(Fig. 7.19) 




Fig. 7. 19 Auxiliary base bisected by the traverse line 



Here 



u b .a 
H = I cot J 

H = — cot — 
1 2 2 

H 2 = y cot T 
H = H + H 2 

= — cot — + cot 

2 L 2 2 J 



TCOtT 
4 2 



cot—- + cot — 
2 2 



If a , /3, and 6 2 are all small, 



H = 


2a 


"2 2 " 




/>[ 


i li 




a ^ 


Ii + K„ 


s/t = 


a 


"i i 



5a 



(7.71) 



(7.72) 



8H 3 = — SB. 
1 aff 



5/Z/9 = 



"ft 



% 



398 



SURVEYING PROBLEMS AND SOLUTIONS 

2 cw, 2 



Total r.m.s. error 8H =y/{Sf£ + 8Hp* + SH/3, 



MfeVl IT 8 A ^ 



If a = /3, = j8 2 and da = 8/3, = S&,, 

a . / 1 „ 2 „2 



8H = 



da 4 8a Sa 
a a a a 



V6 b8a 



(7.73) 



but 



H = 



2b 
2b 



J 2b 
a = J~H 

V6 b// 3/2 5a 



8tf = 



2 3/z b 3/2 

V3 J/ 3 ' 2 8a 

2V^ 



(7.74) 



N.B. This is the same value as for the central auxiliary base (7.68). 

7.65 With two auxiliary bases (Fig. 7.20) 

The auxiliary base f^ is extended twice to H. 




Here 



Fig. 7.20 Two auxiliary bases 

K = —cot — 
6 2 2 

//, = H b cotjS 
H = H x cot<£ 



TACHEOMETRY 
H = — cot— cot /8 cot 



H = -cot/3 cot0 



If a , /3 and are all small, 



H = 


6 
a/80 


SH = 


b§a 
~a 2 /80 


BHp = 


b5/8 
a/3 2 


5fk - 


bd</> 



* a/30 2 
Total r.m.s. error 8H = yJSHj +8H 2 p + SHj 



b l\8a 5/8 2 50' 



a/80/s/ i a 2 /S 2 
If a = /8 = and 5a = 5/8 = 50, 



i.e. 



then 



H b //, H 



b 


H b H, 


», 


= VftH 


H b 


", 2 ,_ 

= — = JUT 

H v ' 


8H 


b /35a 2 
= a 3 */ a 2 




V3 b8a 



399 
(7.75) 

(7.76) 



(7.77) 



but 



8H 



a = 3 Jw 

V3b5aH 4/3 



b 4/3 

4/3, 



V3/^ /3 5a 



J/3 



(7.78) 



400 



SURVEYING PROBLEMS AND SOLUTIONS 



If b= 2m and da = ±1", 
8H = 



8H 
If — = 1/10000, 
H 



and 



y/3H 4 



206 265 x 

H A/3 
150000 



'\/2 



8H 


= 


1 


*JH 


H 


10 000 


150000 


3/ 


= 


15 




H 


= 


3375 metres 


H b 


= 


b_ 
a 

, 1/3 

b 





w 



(7.79) 



fi = * 



a/3 r ,i/3 



H' 



= 2 2/3 H U3 
= 1-5866 x 15 
= 23'8 metres 

7.66 The auxiliary base used in between two traverse lines 

(Fig. 7.21) 




Fig. 7.21 Auxiliary base between two traverse lines 

b a 
H b = -cot- 



H. = 



% sinCft+fl,) 
sin/3, 



TACHEOMETRY 



401 



b a 

2 cot 7j sin (j8, + 0,) 

sin/3, 
bsinQ3 1 + 0,) 



(7.80) 
(7.81) 



Similarly 



«. = 



a sin /3, 

ft a 

2 cot g sin (p 2 + #2) 

sin /3 2 

fcsin(j3 2 +<9 2 ) 

a sin j8 2 

Here the errors are not analysed as the lengths and angles are 
variable. 

Example 7.14 A colliery base line AB is unavoidably situated on 
ground where there are numerous obstructions which prevent direct 
measurement .It was decided to determine the length of AB by the 
method illustrated in Fig. 7.22, where DE is a 50 metre band hung in 
catenary with light targets attached at the zero and 50 metre marks. 

From the approximate angular values shown, determine the maxi- 
mum allowable error in the measurements of the angles such that the 
projection of error due to these measurements does not exceed: 

(a) 1/200000 of the actual length CD when computing CD from 
the length DE and the angle DCE and 

(b) 1/100 000 of the actual length AB when computing AB from 
the angles ACD, CDA, BDC, and DCB and the length DC. 
For this calculation, assume that the length DC is free from 
error. (R.I.C.S.) 

D 50m E 









'QOSvV 








^ 70^~ 


"^70°/ 




A S 








i3>\ 




r 1 




0(J 








V. 70^_ 


■fJO*^ 





c 

Fig. 7.22 



402 



SURVEYING PROBLEMS AND SOLUTIONS 



Assuming Angle DCE(a) = Angle DAB (-|)= 1 / 2 (180-2x70) = 20°, 



(a) 



DC = DE cot a 



(b) 



The error 8DC 


= 


DE 


2 * 
cosec a 8a 




TU ♦• 8DC 
The error ratio 

DC 


= 


DE cosec a 8a 
DE cot a 




= 


8a 1 




sin a cos a 200 000 


8a 


= 


206 265 x Vi sin 2a 
200000 




= 


0-51566 sin 40° = 0-33 seconds 
(say 1/3") 






DC 


\ ft ft] 




AB 


= 


~J 


cot T+ cot-j 





£C 2 ft ^ 

8AB R = ±— cosec y 5/8, 



% 



DC 



ft 



•546^ = ± — cosec —SB. 



Total error 



SAB 



2 
DC 



ft 
cosec 4 — 1 5/3f + cosec 4 — -8B% 



2 



but ft = ft and assuming 5ft = 5ft. 



The error ratio 



8 A B = ± — V 2 cosec 2 — 5/8 

SAB DC/2 y/2 cosec 2 B/28B 
~AB = DC/2 x 2 cot 0/2 

V2 5/3 

= 2 sin ft/2 cos jS/2 

V25/3 _ 1 
= sin/6 = 100000 

206 265 sin 40° 



5/3 = 



V2x 100000 
= 0*94 seconds (say ± 1") 



TACHEOMETRY 



403 



Exercises 7(b) 

11. (i) What do you understand by systematic and accidental errors 
in linear measurement, and how do they affect the assessment of the 
probable error? 

Does the error in the measurement of a particular distance vary in 
proportion to the distance or to the square loot of the distance? 

(ii) Assume you have a subtense bar the length of which is known 
to be exactly 2 metres (6*562 ft) and a theodolite with which horizontal 
angles can be measured to within a second of arc. In measuring a 
length of 2000 ft., what error in distance would you get from an angular 
error of 1 second? 

(iii) With the same equipment, how would you measure the dis- 
tance of 2000 ft in order to achieve an accuracy of about 1/5000? 

(Aide memoire: 1 second of arc = 1/206265 radians.) 

(I.C.E. Ans. (ii) ±2-95 ft ) 

12. (a) When traversing with a 2 metre subtense bar, discuss the 
methods which can be adopted to measure lines of varying length. In- 
clude comments on the relative methods of angular measurement by 
repetition and reiteration. 

(b) A bay length AB is measured with a subtense bar 2 metres 
in length, approximately midway between and in line with AB. 
The mean angle subtended at A = 1°27'00" 

at B = 1°35'00" 
Calculate the length AB. 

(E.M.E.U. Ans. 151-393 m) 

13. The base AB is to be measured using a subtense bar of length 
b and the double extension layout shown in the figure. 

If the standard error of each of the two measured angles is + 8a 
develop a formula for the proportional standard error of the base length. 

Find the ratio a^ : a 2 which will give the minimum proportional 
standard error of the base length. 

What assumptions have you made in arriving at your answers? 




(R.I. C.S.) 



404 



SURVEYING PROBLEMS AND SOLUTIONS 



14. (a) Describe, with the aid of sketches, the principles of sub- 
tense bar tacheometry. 

(b) The sketch shows two adjacent lines of a traverse AB and 
BD with a common sub-base BC. 

Calculate the lengths of the traverse lines from the following data: 

Angles BAC = 5°10'30" 

CBA = 68°56'10" 

YBX = 1°56'00" 

CDB = 12°54'20" 

DBC = 73°18'40" 

Length of bar 2 metres. 

(E.M.E.U. Ans. AB, 631-96 m; BD, 264-78 m) 




vi Y Not to scale 

X C Y 
Fig. 7.24 

15. Describe the 'single bay* and 'double bay' methods of measuring 
linear distance by use of the subtense bar. 
Show that for a subtense bar 



L = — cot cf>/2 

where L = the horizontal distance between stations, 

S = length of subtense bar, 

<f> = angle subtended by targets at the theodolite. 

Thereafter show that if S = 2 metres and an error of ± A<£ is made 
in the measurement of the subtended angle, then 

AL A0L 

L = ~ 2 

where AL is the corresponding error in the computed length. 

Assuming an error of ± lsec(A<£) in the measurement of the sub- 
tended angle what will be the fractional error at the following lengths? 

(a) 50 metres (b) 100 metres (c) 500 metres. 

(N.U. Ans. (a) 1/8250; (b) 1/4125; (c) 1/825.) 



TACHEOMETRY 



405 



Exercises 7(c) (General) 

16. The following readings were taken with a theodolite set over a 
station A, on to a staff held vertically on two points B and C. 

Inst. St. Horizontal Circle Vertical Circle Stadia Readings Staff 

Reading Reading U M L St. 

A 33°59'55" +10°48' 8-44 6-25 4-06 B 

A 209°55'21" - 4°05' 7-78 6-95 6-12 C 

If the instrumental constant is 100 and there is no additive con- 
stant; calculate the horizontal distance BC and the difference in 
elevation between B and C. (E.M.E.U. Ans. 587*48 ft; 93*11 ft) 



17. Readings were taken on a vertical staff held at points A, B, and 
C with a tacheometer whose constants are 100 and 0. If the horizon- 
tal distances from instrument to staff were respectively 153, 212, and 
298 ft, and the vertical angles +5°, +6° and -5°, calculate the staff 
intercepts. If the middle-hair reading was 7*00 ft in each case what 
was the difference in level between A, B and C? 

(L.U. Ans. 7-77/7-00/6-23; 8-07/7-00/5-93; 8-50/7-00/5-50; A-B. 

+ 8-88; B-C. -48-30) 

18. A theodolite has a tacheometric multiplying constant of 100 and 
an additive constant of zero. When set 4*50 ft above a station B, the 
following readings were obtained: 



Station 
at 



Sight 



Horizontal 
Circle 



Vertical 
Circle 



Stadia Readings 



Top 



Middle 



Bottom 



A 
C 



028°21'00" 
082°03'00" 



20°30' 



3-80 



7-64 



11-40 



The co-ordinates of station A are E 546-2, N 0-0 and those of 
B are E 546-2 N -394-7. 

Find the co-ordinates of C and its height above datum, if the 
height of station B above datum is 91-01 ft. 

(L.U. Ans. 1083-6 E 0-1 N; +337-17 ft) 



19. The following readings were obtained in a survey with a level 
fitted with tacheometric webs, the constant multiplier being 100 and 
the additive constant zero. 



406 



SURVEYING PROBLEMS AND SOLUTIONS 



Inst, at 


Point 


Staff Readings 


A 


B.M. 207*56 


1-32 


2-64 


3-96 




B 


2-37 


3-81 


5-25 


C 


B 


5-84 


7-95 


10-06 




D 


10-11 


11-71 


13-31 




E 


8-75 


9-80 


10-85 


F 


E 


11-16 


13-17 


15-18 




T.B.M. 


3-78 


5-34 


6-90 



Subsequently the level was tested and the following readings ob- 
tained: 



t. at 


Point 


Staff Readings 


P 


Q 


4-61 


5-36 6-11 




R 


3-16 


3-91 4-66 


S 


Q 


4-12 


4-95 5-78 




R 


3-09 


3-17 3-25 



Find the level of the T.B.M. 

(L.U. Ans. 212-98 ft) 

20. A theodolite was set up at P, the end of a survey line on uni- 
formly sloping ground and the readings taken at approximately 100 ft 
intervals along the line as follows: 



At 


Point 


Elevation Angle 


Stadia Readings 


P 


A 


4°16' 


3-66 


4-16 


4-66 




B 


4° 16' 


2-45 


3-46 


4-47 




C 


5°06' 


1-30 


2-82 


4-34 




D 


5°06' 


5-87 


7-88 


9-89 




E 


5°06' 


6-15 


8-65 


11-15 



An error of booking was apparent when reducing the observations. 
Find this error, the levels of the points ABCDE and the gradient 
PE, if the ground level below the instrument was 104-20 O.D. and 
the height of the instrument 4-75. Instrument constants 100 and 0. 
(L.U. Ans. A 112-21, B 120-48, C 128-68, D 136-66, E 144-57; 

Grad. 1 in 12-28) 

21. The following data were taken during a survey when stadia read- 
ings were taken. The levelling staff was held vertically on the stations. 
The height above datum of station A is 475-5 ft above Ordinance Dat- 
um. The multiplying factor of the instrument is 99-5 and the additive 
constant 1*3 ft. Assume station A to be the point of origin and cal- 
culate the level above Ordinance Datum of each station and the horiz- 
ontal distance of each line. 



TACHEOMETRY 



407 



Back 


Instrument 


Fore 


Instrument 


Horizontal 


Vertical 


Staff Readings 


Station 


Station 


Station 


Height 


Angle 


Angle 


Upper Middle Lower 


- 


A 


B 


4-95 


- 


+ 5°40' 


9.90 


8*00 


6.10 


A 


B 


C 


5-00 


164°55' 


+ 7°00' 


8.44 


6.61 


4.7.8 


B 


C 


D 


5-10 


179°50' 


-8°20' 


9.20 


7.57 


5-94 



(R.I.C.S. Ans. A 475-50, B 509-73, C 552-33, D 502-66; AB 375-70, 

BC 360-04, CD 322-23) 

22. The undermentioned readings were taken with a fixed-hair tach- 
eometer theodolite on a vertical staff. The instrument constant is 100. 
Calculate the horizontal distance and difference in elevation between 
the two staves. 



Instrument 


Horizontal 


Vertical 




Staff Station 


Station 


Circle 


Circle 






X 


33°59'55" 


+ 10°48'00" 


[8-44 , | 

6-25} 

1 4-06 J 


A 


X 


209°55'21" 


- 4°05'00" 


(7-78] 

6-95 

U-12J 


B 



(MQB/S Ans. 587-5 ft; 93-1 ft) 

23. The undernoted readings were taken at the commencement of a 
tacheometric survey, the multiplying factor of the tacheometer being 
100 and the additive constant 1-3 ft. 

Calculate the co-ordinates and reduced level of station D assum- 
ing A to be the point of origin and the reduced level there at 657*6 ft 
above datum. The azimuth of the line AB is 205° 10' . 

(MQB/S Ans. S 893-83 ft; W 469-90 ft; 718-54 ft) 

24. The following readings were taken by a theodolite used for tach- 
eometry from a station B on to stations A, C and D: 



Sight 



A 
C 
D 



Horizontal 
Angle 



301°10' 
152° 36' 
205°06' 



Vertical 
Angle 



Stadia Readings 
Top Centre Bottom 



-5°00' 3-48 7-61 11-74 
2°30' 2-15 7-92 13-70 

The line BA has a bearing of N 28°46' E and the instrument 
has a constant multiplier of 100 and an additive constant zero. Find 
the slope of the line CD and its quadrantal bearing. 

(L.U. Ans. 1 in 7*57; N 22° 27' 10" W) 



408 SURVEYING PROBLEMS AND SOLUTIONS 

25. The following observations were taken with a tacheometer (mult- 
iplier 100, additive constant 0) from a point A, to B and C. 

The distance EC was measured as 157 ft. Assuming the ground to 
be a plane within the triangle ABC, calculate the volume of filling 
required to make the area level with the highest point, assuming the 
sides to be supported by concrete walls. 

Height of instrument 4'7ft, staff held vertically. 

At To Staff Readings Vertical Angle 

A B 1-48 2-73 3-98 +7°36' 

C 2-08 2-82 3-56 -5°24' 

(L.U. Ans. 297600 ft 3 ) 

26. Explain the principles underlying the measurement of a horizontal 
distance by means of stadia readings. 

Using stadia readings, the horizontal distance between two stat- 
ions A and B is found to be 301*7 ft. 

The difference in height between the two stations is 3*17 ft. Cal- 
culate the appropriate stadia readings, stating clearly the assumptions 

you have made. 

(R.I.C.S.) 

27. (i) Derive expressions for the probable errors of determination of 
horizontal and vertical distances tacheometrically due to known probab- 
le errors ds in the measurement of stadia intercept and da in the meas- 
urement of the vertical angle. 

It may be assumed that an anallactic instrument is used and the 
staff is held vertically. 

(ii) In a series of tacheometric observations a sight is taken to 
the top of a building where the vertical angle is about 11°30'. If the 
stadia intercept is 2-52 ft and ds = +0-002 5 ft and da = ± 1' what are 
the probable errors in determining the horizontal distances? The tach- 
eometric constant is 100. 

(iii) Assuming that using a staff graduated to 0'Olft each stadia 
hair can be read correctly to the nearest 0*01 ft will such a staff be 
good enough to give an accuracy of 1 : 500 over distances from 100 to 
500 ft, and if not what accuracy can be achieved? It may be assumed 
that the vertical angles will be small and errors in the vertical angle 
can be ignored. 

(R.I.C.S. Ans. (ii) ±0-24 ft; ±0-03 ft) 

28. A third order traverse line AB is measured by the following 
method: measure angles a,, a 2 , 6 shown on the diagram; measure dis- 
tances AC,, AC Z , by the angles at A to a subtense bar at C, and C 2 . 

Two measures of AB are thus obtained, their mean being accept- 
ed. 



TACHEOMETRY 409 




A 2m subtense bar is centred at C, and C 2 and oriented at right 
angles to AC X and AC 2 . 

Observed horizontal angles are as follows. 
Subtense angles at A: to C, = 1°05'27-1" 
to C 2 = l°04'30-0" 
6 = 100°35'33" a, = 2°51'27"; a 2 = 2°53'55" 
Compute the horizontal distance AB. 

(R.I.C.S. Ans. 104-70 m) 

29. Describe in detail how you would determine the tacheometric con- 
stants of a theodolite in the field. Show how the most probable values 
could be derived by the method of least squares. 

Sighted horizontally a tacheometer reads r, = 6*71 and r 3 = 8*71 
on a vertical staff 361*25 ft away. The focal length of the object glass 
is 9 in and the distance from the object glass to the trunnion axis 6 in. 
Calculate the stadia interval. 

Given D = — s + (/+ c) 
i 

(N.U. Ans. 0-05 in.) 

30. Describe the essential features of a subtense bar and show how 
it is used in the determination of distance by a single measurement. 
Allowing for a 1 second error in the measurement of the angle, cal- 
culate from first principles the accuracy of the measurement of 200 ft 
if a 2 metre subtense bar is used. Show how the accuracy of such a 
measurement varies with distance and outline the method by which max 
imum accuracy will be obtained if subtense tacheometry is used in the 
determination of the distance between points situated on opposite banks 
of a river about 600 ft wide. 

(I.C.E. Ans. 1 in 6 800) 

31. An area of ground was surveyed with a fixed stadia hair tacheo- 
meter (constants 100 and 0) set up in turn at each of four stations 

A, B, C and D. Observations were made with the staff held vertically. 
A BCD A formed a closed traverse and it was found that the difference 
in level between these four stations as calculated from the tacheomet- 
ric readings would not balance. 



410 



SURVEYING PROBLEMS AND SOLUTIONS 



Later it was realised that a new altitude bubble had recently been 
fitted to the instrument but unfortunately had not been correctly adjust- 
ed. In order to determine the true differences of level between the four 
stations, a level known to be in perfect order was used. 

Fieldbook observations from both surveys are as follows: 



Instrument 


Height of 


Staff 


Vertical 


Stadia Readings 


Station 


Instrument 
(ft) 


Station 


Angle 


(ft) 


A 


4-55 


B 


+ 3° 


4-05 5-66 7-27 


B 


4-60 


C 


+ 2°20' 


8-71 11-02 13-33 


C 


4-70 


D 


-2°30' 


3-74 5-67 7-60 


D 


4-50 


A 





2-38 4-99 7-60 



Backsight 


Intermediate sight 


Foresight 


Remarks 


11-78 


- 


- 


Station A 


10-30 


- 


5-04 


Change point 


- 


3-19 


- 


Station B 


9-84 


- 


1-65 


Change point 


3-27 


- 


1-69 


Station C 


2-83 


- 


14-78 


Change point 


- 


11-34 


- 


Station D 


6-41 


- 


9-70 


Change point 


- 


- 


11-57 


Station A 



Calculate the vertical angles that would have been observed from 
stations A and C if the altitude bubble of the tacheometer had been 
in correct adjustment. 

Describe the procedure which should be adopted in correcting the 
adjustment of the altitude bubble, identifying the type of instrument 
for which your procedure is appropriate. 

(I.C.E. Ans. +2°40'; -2°50') 



Bibliography 

THOMAS, W.N., Surveying (Edward Arnold). 

BANNISTER, A. and RAYMOND, S., Surveying (Pitman). 

REDMOND, F.A., Tacheometry (Technical Press). 

HIGGINS, A.L., Higher Surveying (Macmillan). 

HODGES, D.J., Articles in Colliery Engineering, Dec. 1965 - Oct. 1966. 

GEISLER, M., 'Precision tacheometry with vertical subtense bar applied 

to topographical surveying', Survey Review, Oct. 1964. 
WATSON and SHADBOLT, 'The subtense bar applied to mine surveying', 

Chartered Surveyor, Vol. 90, No. 11, May 1958. 



8 DIP AND FAULT PROBLEMS 



Problems on gradients take a number of different forms and may be 
solved graphically or trigonometrically according to the accuracy re- 
quired. 

8.1 Definitions 

Let ABDE represent a plane inclined to the horizontal at a°, 
Fig. 8.1. 

Level line (strike) D 




S JC 



A B 

Fig. 8.1 

Dip. The dip of a bed, seam or road in any direction is the angle of 
inclination from the horizontal plane. 
It may be expressed as: 

(a) An angle from the horizontal, e.g. 6°03' , 

(b) A gradient, 1 vertical in n horizontal (the fraction 1/n repre- 
sents the tangent of the angle of inclination, whilst n represents 
its cotangent) or 

(c) A vertical fall of x inches per horizontal yard, e.g. 3 inches 
per yard. 

N.B. The term rise denotes the opposite of dip. 

Full Dip (or true dip) is the maximum inclination of any plane from 
the horizontal and its direction is always at right-angles to the mini- 
mum inclination (i.e. nil) or level line known as strike. 

In Fig. 8.1, lines AE and BD are lines of full dip, whilst ED 
and AB are level lines or strike lines. 

Apparent Dip is the dip observed in any other direction. It is always 
less than full dip and more than strike. In Fig. 8.1 the line AD is an 

411 



412 



SURVEYING PROBLEMS AND SOLUTIONS 



apparent dip inclined at. an angle of /8° in a direction 6° from full dip. 

Depth of Strata. The depth of a stratum is generally measured relative 
to the surface, to Ordnance Mean Sea Level (as used in levelling) or, in 
order to obtain positive values, may be expressed relative to some 
arbitrary datum, e.g. the N.C.B. (National Coal Board) datum, which 
is 10 000 ft below M.S.L. 

Thickness of Strata. The true thickness is measured at right- angles 
to the bedding plane. For inclined strata penetrated by vertical bore- 
holes, an apparent thickness would be derived from the borehole core. 

Example 8.1 A vertical borehole passes through a seam which is 
known to dip at 40°. If the apparent thickness of the seam as shown by 
the borehole core is 5 ft calculate: (a) the true thickness of the seam; 
(b) the gradient of the seam. 




Fig. 8.2 

(a) True thickness t = 5 cos 40° = 3*83 ft 

= 3 ft 10 in. 

(b) Gradient of seam cot 40? = M92 
••• Gradient is 1 in 1*192 



Example 8.2 If a seam dips at 1 in 4 what is the true area of one 

square mile in plan 

Cot0 = 4 .*. 6 = 14° 02' 

True length of one dipping side 

1760 

~ COS0 



DIP AND FAULT PROBLEMS 413 

1760 



0-970 15 
True area 



yd 



= 176 ° 2 = 3192908 sq. yd 
0-97015 ±JL ~ 



compared with 3097 600 sq. yd in plan. 

8.2 Dip Problems 

8.21 Given the rate and direction of full dip, to find the apparent dip 
in any other direction 

Let Fig. 8.3 represent the plan. 
Graphical Solution 

Draw AB and AC representing 

strike and full dip. Let the length 

AC be x units long. As the line AC £ 

dips at I'm x, C will be 1 unit 

vertically below A. Draw the strike 

line CE through C, which is now 1 

unit below strike line AB. Fig. 8.3 

Any line starting from A will then be 1 unit vertically below A when it 

it cuts the line CE, and its gradient will be 1 in y, where y is the 

length measured in the same units. 

Trigonometrical solution 

Triangle ADC is right-angled at C 

AC 



A 


Strike 




B 


>< 




^^■7/ 






c 


" e 


>s^> 






5 
(n 


3' 

X 






E 


C 

• 


' 




D 





AD = 



COS0 



ie - y = ^k?) g- 1 ) 

COSC7 

or x - y cos# (8.2) 

i.e. the gradient value of full dip x = the gradient value of apparent 
dip y x cosine of the angle between. 

8.22 Given the direction of full dip and the rate and direction of an 
apparent dip, to find the rate of full dip (Fig. 8.4) 

Graphical solution 

Draw directions AB and AC of full dip and apparent dip respec- 
tively. 

Let AC = y units. 

Draw CD perpendicular to AB through C cutting the full dip 



414 



SURVEYING PROBLEMS AND SOLUTIONS 



direction at B. 

Length AB = x units is the gradient equivalent of full dip. 




Fig. 8.4 
Trigonometrical solution 

AB = AC cos0 

i.e. x - ycos(/3-a) (8.3) 

i.e. cot full dip = cot apparent dip cosine angle between (8.4) 

or tan apparent dip = tan full dip cosine angle between (8.5) 

Example 8.3 The full dip of a seam is 1 in 4 N 30° E. Find the 
gradients of roadways driven in the seam 

(a) due N, (b) N 75° E, (c) due E. 





<o 










c 








c 










HI 

"c 
" — *> 


>/ 




^i. 


— ^\° 6 


^v£ 




--*~ 8 units 




_ ^ Due E 










f^\1 in 8 






N^ 







Graphically, 



Fig. 8.5 

(a) AC due N 1 in 4*6 

(b) AD N 75° E 1 in 5*7 

(c) AE due E 1 in 8 



DIP AND FAULT PROBLEMS 



415 



Trigonometrically, 
(a) AC = 



(b) AD = 



cos 30° 0-8660 



cos 45° 0-7071 



= 4-618 (12° 13' ) 



= 5-657 (10° 02' ) 



^ AE = OTo = «£ =8-0 (7" 08') 

N.B. (a) Lines at 45° to full dip have gradients 1 in yJ2x, 
e.g. 1 in y/2 x 4 = 1 in 5-657 
(b) Lines at 60° to full dip have double the gradient value, 
i.e. 1 in 2x, 

e.g. 1 in 2x4 = 1 in 8 

8.23 Given the rate and direction of full dip, to find the bearing of 
an apparent dip (Fig. 8.6) 




Fig. 8.6 

This is the converse of 8.22 but it should be noted that there are 
two directions in which a given apparent dip occurs. 

Graphical solution 

Plot full dip, direction a°, of length x units as before. 
Draw strike lines through A and B. 

With centre A draw arcs of length y to cut strike line through B, 
giving 6° on either side of AB as at AC and AD. 

Trigonometrical solution 

a x 
cosO = — 

y 



416 



SURVEYING PROBLEMS AND SOLUTIONS 



Bearing AC = a-d 

AD = a+0 

Example 8.4 A seam dips 1 in 5 in a direction 208° 00' . In what 
direction will a gradient of 1 in 8 occur? 




By plotting AC 156*45' 
AD 259° 15' 



Trigonometrical ly 



.". Bearing AC 
AD 



Fig. 8.7 

6 = cos -1 5/8 

= 51° 19' 

= 208° 00' - 51° 19' = 156° 41' 

= 208° 00' + 51° 19' = 259° 19' 



8.24 Given two apparent dips, to find the rate and direction of full 
dip (Fig. 8.8) 




Fig. 8.8 



DIP AND FAULT PROBLEMS 



417 



Graphical Solution 

Plot direction of apparent dips AC and AD of length y and z 
units respectively. 

Join CD. 

Draw AB perpendicular to CD. Measure AB in the same units as 
y and z. 
Trigonometrical solution 




Fig. 8.9 



In triangle ADC, Fig. 8.9, AC = y AD = z DAC = 6 

C ~D z-y tan 180 - $ 

Tan — - — = — = - 

2 z + y 2 

From this, angles C and are known and thus A. 

Triangle ABC may now be solved 

AB = AC cos A 

Bearing AB = Bearing AC - A 



Alternative solution 
also 



x = y cos A 
x = z cos(0 - A) 
y cos A = z cos(0 - A) 



Dividing by cos A 



= z (cos 6 cos A + sin 6 sin A) 
y = z (cos $ + sin 6 tan A) 



418 SURVEYING PROBLEMS AND SOLUTIONS 

Thus tanA = — 1 22£* 

z sin0 sin0 

= ycosecfl _ cQte 

z 

If y and z are given as angles of inclination, 

y = cot a 
z = cot )6 

then tanA = tan/3 cosectf cot a- cot0 (8.6) 

This gives the direction of full dip. 

The amount x = cot a cos A 

or cotS = cotacosA (8.7) 

or tana = tan 8 cos A (8.8) 

N.B. If 6 = 90°, by Eq.(8.6), 

tanA = tan/3 cot a (8.9) 

Alternative solution given angles of inclination, Fig. 8.10 

a. 



Fig. 8.10 

Let A, B and C be three points in a plane with B at the highest 
level and A at the lowest. 

Let AB } C^D be a horizontal plane through A, with B, and C, 
vertically below B and C respectively and D the intersection on this 
plane of BC and B, C, each produced . 

AB^C^ represents the plan of points A,B and C. 

In the plane ABCD, AD is in the horizontal plane and is therefore 
a line of strike. BE is perpendicular to AD and is therefore the line 
of full dip. 

In the right-angled triangle ABB V BB, = AB, tana. 

In the right-angled triangle DB, B, BB, = DB, tan/3 

AB y tana = DB y tan/3 



DIP AND FAULT PROBLEMS 



419 



AB. 



i.e. 



DB, 
Also in triangle AB X D 

AB, 
DB. 



tanjS 
tana 



sine 



K 



but 



sin^zS 

sine _ tan/6 _ 
sin<£ tana 

e + 0= 180 - 6 = P 

e = P-cf> 

sin(P - cf>) = 
sin<£ 



i.e. 



K sin<£ 



cot<£ 



(8.10) 



= sin P cos (f> - cos P sin <f> 

= sin P cot <f> - cos P 

= sin(180 - 0) cot <£ - cos(180 - 0) 

= sin cot + cos 

K-cosO tanfl . 

— z — = t . ~ - cot0 

sine/ tan a sin 

The value of <£ will then give the direction of the strike with full 
dip at 90° to this. 

The inclination of full dip (8) is now required. 

Let AB t = cot a 
then B, E = cot 5 

as before therefore cot 8 = cot a sin 4> (8.11) 

i.e. Eq.(8.7) cotS = cotacosA. (A = 90 - <£) 

Example 8.5 A roadway dips 1 in 4 in a direction 085° 30' , inter- 
sects another dipping 1 in 6, 354° 30'. Find the rate and direction of 
full dip. 



Fig. 8.11 




420 SURVEYING PROBLEMS AND SOLUTIONS 



Graphically Full dip 051° 00' 1 in 3*3 

Trigonometrically 

C -D 6 - 4 . 180 - (360° - 354° 30' + 085° 30' ) 

tan = tan 7z 

2 6+4 2 

2^ tan 89° 00' 
"10 2 

= I tan 44° 30' 
C ' D = 11° 07' 10" 



2 

n = 44° 30' 00" 



C + D Mom'nn" 



2 
C = 55° 37' 10" 

A = 34° 22' 50" 

Bearing of full 

dip = 085° 30' - (90° - 55° 37' 10") 

= 051° 07' 10" 
(AB) = 4 sin 55° 37' 10" 
= 4x0-825 31 
= 3-30124 
Gradient of full dip = 1 in 3*3 
Alternative solution 

Gradient 1 in 4 = 14° 02' 10" 
Gradient 1 in 6 = 9° 27' 40" 
From Eq. (8.6), 

* x _ tan9°27'40 w _ cot91 o 00 * 

tanA " tanl4°02'10"sin91°00" 

0-166 64 nni t^ 

= 0-25000 x 0-99985 + °-° 1746 
= 0-666 66 + 0-017 46 
= 0-68413 
A = 34° 22' 40" 

Bearing of full dip = 085° 30' 00" - 34° 22' 40" 
= 051° 07' 20" 

Rate of full dip 

x = cot 8 = cot 14° 02' 10" cos 34° 22' 40" 
= 4 x 0-825 33 



DIP AND FAULT PROBLEMS 



421 



= 3-301 (Gradient 1 in 3*3) 
8 = 16° 51' 10" 



8.25 Given the rate of full dip and the rate and direction of an appar- 
ent dip, to find the direction of full dip (Fig. 8.12) 




Fig. 8.12 

There are two possible solutions. 

Graphical solution 

Draw apparent dip AC in direction and of y units. 
Bisect AC and draw arcs of radius l / 2 y. 
Through A draw arcs of length x to cut circle at B, and B 2 
This gives the two possible solutions /IS, and AB Z 

Trigonometrical solution 
In triangle AB } 

AB, = x 

AO = OB, = l / 2 y 



1 = /(s-x)(s-y 2 v) = f 

2 V s(s-y 2 y) V 



where s = y 2 [ x + y 2 y + l / 2 y] = l / 2 (x + y) 

2 V {x + y) 
Bearing of full dip = £ ± 6 



(8.12) 



422 



SURVEYING PROBLEMS AND SOLUTIONS 



Example 8.6 A roadway in a seam of coal dips at 1 in 8, 125° 30', 
Full dip is known to be 1 in 3. Find its direction. 




ByEq.(8.12), tan 



■/ 



Fig. 8.13 



(8-3) 



2 aJ (8 + 3) 
= 0-674 20 



5_ 
11 



e 



= 33° 59' 20" 



6 = 67° 58' 40" 



Bearing of full dip 

= 125° 30' 00" 
= 57° 31' 20" 



67° 58' 40" 



or = 125° 30' 00" + 67° 58' 40" 
= 193° 28' 40" 

8.26 Given the levels and relative positions of three points in a plane 
(bed or seam) , to find the direction and rate of full dip 

This type of problem is similar to 8.24 but the apparent dips have 
to be obtained from the information given. 

To illustrate the methods 

Draw an equilateral triangle ABC pf sides 600 ft each. 

Example 8.7 If the levels of A, B and C are 200, 300 and 275 ft 
respectively, find the rate and direction of full dip. 



DIP AND FAULT PROBLEMS 



423 




Fig. 8.14 

Select the highest point, i.e. B — gradients will then dip away from B. 

Semi-graphic method 

As B is the highest point and A is the lowest the level of C 

must be between them. 

Difference in level AB = 300 - 200 = 100 ft 

Difference in level CB = 300 - 275 = 25 ft 

25 
.*. Level of C must be -r^r- of distance AB from B. 

.'. Gradient of full dip = 25 ft in 144 ft (scaled value) 

= 1 in 5-76 
Direction scaled from plan 14° E of line AB. 

N.B. Strike or contour lines in the plane may be drawn parallel as 
shown. 



Alternative method 

Gradient B - A 

i.e. 
Gradient B-C 



= (300 - 200) ft in 600 ft 

= 1 in 6 

= (300 - 275) ft in 600 ft 

= 1 in 24. 



Lay off units of 6 and 24 as in previous method to give x and y, 
Fig. 8. 15. Line XY is now the strike line and BZ perpendicular to 
XY produced is the direction of full dip. Length BZ represents the 
relative full dip gradient value. 



424 



SURVEYING PROBLEMS AND SOLUTIONS 



S*£i 





^-14° 


/ ^ — ■—" 6 units * 


A 5-8 units 


c - """x 24 units 


fl 


Fig. 8.15 




By calculation: 




In triangle XYB, 




tan V = 24 + 6 tan60 ° 




_ 18x^732 . ,.0392 




Y ~ X = 43° 54' 




Y t X - 60°00' 





Y = 103° 54' 
In triangle YZB, 

ZB = 6 sin 103° 54' 

= 6 x 0-970 72 = 5-82 (Gradient of full dip). 
Angle YZB = 103° 54'- 90° = 13° 54' E of line AB. 

Example 8.8 From the following information, as proved by boreholes, 
calculate the direction and rate of full dip of a seam, assuming it to be 
regular. 



Borehole 



Level of surface 
above Ordnance Datum (ft) 



Depth from surface to 
floor of seam (ft) 



370 
225 
255 



1050 
405 
185 



Borehole B is 1414 yards from A in the direction 340° 
Borehole C is 1750 yards from A in the direction 264° 



DIP AND FAULT PROBLEMS 



425 




Fig. 8.16 



Gradient 



At 


A, Surface 


+ 


370 




Depth 


- 


1050 




Seam level 


- 


680 ft 


At 


B, Surface 


+ 


225 




Depth 


- 


405 




Seam level 


— 


180 ft 


At 


C, Surface 


+ 


225 




Depth 


- 


185 




Seam level 


+ 


70 ft 



CA = (680 + 70) ft in 3 x 1750 ft i.e. 1 in 7. 

BA = (680 - 180) ft in 3 x 1414 ft i.e. 1 in 8-484 
.-. Let C,4 = 7 units and B^A = 8*484 units. 
In triangle AB % C, 

Angle A = 340 - 264 = 76° 

2 



tan 



8-484 - 7 180 - 7 6 
8-484 + 7 tan 2 
1-484 tan 52° 



15-484 



= 0-12267 



Cl I gl = 7° 00', Cl + Bl = 52° 00' 
2 2 

••• C, = 59° 00' 



426 



SURVEYING PROBLEMS AND SOLUTIONS 



In triangle AC^X, XA is full dip perpendicular to C, B, 

••• XA = C,A sinC, 

= 7 sin59° = 6 



full dip is 1 in 6. 

Bearing AC, = 264° 
Bearing AX = 264° + (90 - 59) = 295 c 
Bearing of full dip is XA = 115° 



Example 8.9 Three boreholes A, B and C are put down to prove a 
coal seam. The depths from a level surface are 735 ft, 1050 ft, and 
900 ft respectively. The line AB is N 10° E a distance of 1200 ft, 
whilst AC is N55°W, 900 ft. 

Calculate the amount and direction of full dip. 



B is the lowest and A is the highest. 

6 = 55 + 10 = 65° 

1200 1200 

1050 - 735 315 

= 3-809 52 
a = 14° 42' 30" 

r _ 900 _ 900 

COt P ~ 900 - 735 ~ 165 

= 5-45454 
jS = 10° 23' 20" 




4-735 



Fig. 8.17 



Then by Eq.(8.10) 
cot<£ = 



tanjS 



tana sin0 
0-18333 



- cot0 



= 0-26250 x 0-906 31 "°' 46631 
= 0-770 62-0-466 31 =0-304 31 
<f> = 73° 04' 30" 



Bearing of full dip = 010° - (90 - 73° 04' 30") 

= 353° 04' 30" = N6°55'30"W 



DIP AND FAULT PROBLEMS 



427 



Amount of full dip cot 8 = cot a sin ^ 

= 3-809 52 x 0-95669 

= 3-644 53 
i.e. Gradient 1 in 3*64 
Inclination (8) = 15° 20' 40" 

8.3 Problems in which the Inclinations are Expressed 
as Angles and a Graphical Solution is Required 

To illustrate the processes graphical solutions of the previous 
examples are given. 

8.31 Given the inclination and direction of full dip, to find the rate 
of apparent dip in a given direction 

Example 8.10 Full dip N 30° E 1 in 4 (14° 02') 

Apparent dip (a) Due N, (b) N 75° E. 



Apparent dips 

(a) 12*15' 

(b) 10*00' 




10*00' 



Fig. 8.18 

Draw AB (full dip) N 30° E of convenient length say 3 in. 
Through A and B draw strike lines AX and BY and assume that AX 
is 1 unit vertically above BY. At B set off the inclination of full dip 
(i.e. 14° 02') to cut AX at B,. AB^B may now be considered as a 
vertical section with AB A of length 1 unit. Draw a circle of centre A, 
radius AB^ (1 unit). Now draw the direction of apparent dips 

AC (due N) to cut strike BY at C 

and AD (N 75° E) to cut strike BY at D. 



428 



SURVEYING PROBLEMS AND SOLUTIONS 



Also draw AC, and AD, perpendicular to AC and AD respectfully, 
cutting the circle at C, and D,. 
N.B. AB, = AC, = AD, =1 unit. 

AC 
Then -^ represents the gradient of AC (1 in 4*6). The angle 

ACC, is the angle of dip 12° 15'. 

AD 
Similarly -^j- represents the gradient of AD (1 in 5-7). The angle 

ADD, is the angle of dip 10°00' . 

8.32 Given the inclination and direction of full dip, to find the direc- 
tion of a given apparent dip 

Example 8.11 Full dip 1 in 5 (11°18') 208°00'. 
Apparent dip 1 in 8 (7°07'). 



90*- 



Direction of 
apparent dip 
259° 30' and 156*30' 




Fig. 8.19 

Draw AB in direction as before of say 3 in. At A and B draw the 
strike lines AX and BY. At B set off the angle of full dip 11°18' 
to cut AX produced at A, . Draw a circle of centre A and radius AA^ 

Produce BA to cut the circle at Q and set off the angle (90° - 
7°07 ; ) to cut AX at P— this represents a section of the apparent dip, 
AP being the length of the section proportional to the vertical fall of 
1 unit G4Q). With centre A and radius AP draw an arc to cut BY at 
C, and C 2 , i.e. AC, = AC Z = AP. These represent the direction of 
the apparent dips required. 



DIP AND FAULT PROBLEMS 



429 



8.33 Given the inclination and direction of two apparent dips to find 
the inclination and direction of full dip 



Example 8.12 



1 in 4 (14°) 085° 30' 
1 in 6 (9°30') 354° 30' 



Direction of full dip 051* 00"! 
Amount " " " 17' 




Fig. 8.20 

Draw AC and AD in the direction of the apparent dips. With A as 
centre draw a circle of unit radius. Draw AC^ and AD^ perpendicular 
to AC and AD respectively. Set off at C, (90° - 14°) and at D 
(90° - 9°30'). This will give 14° at C and 9°30' at D. Join CD, the 
strike line. Draw AB perpendicular to CD through A and AB^ perpen- 
dicular to AB. Join B, B and measure the direction of AB and the 
angle of inclination B^BA. 



Graphical solution: 

Full dip 17 c 



051°00'. 



Exercises 8(a) 

1. The full dip of a seam is 4 inches in the yard. Calculate the 
angle included between full dip and an apparent dip of 3 inches in the 
yard. 

(Ans. 41° 24') 

2. The angle included between the directions of full dip and apparent 
dip is 60°. If the apparent dip is 9°10', calculate the full dip, expres- 
sing the answer in angular measure, and also as a gradient. 

(Ans. 17° 50'; 1 in 3*1) 

3. On a hill sloping at 18° runs a track at an angle of 50° with the 
line of greatest slope. Calculate the inclination of the track and also 
its length, if the height of the hill is 1500 ft. 

(Ans. 11° 48'; 7335 ft) 



430 



SURVEYING PROBLEMS AND SOLUTIONS 



4. The full dip of a seam is 1 in 3 in a direction N 85° 14' W. A road- 
way is to be driven in the seam in a southerly direction dipping at 1 in 
10. Calculate the quadrant and azimuth bearings of the roadway. 

(Ans. S 22°14'W; 202° 14') 

5. The full dip of a seam is 1 in 5, N 4° W. A roadway is to be set 
off rising at 1 in 8. Calculate the alternative quadrant bearings of the 
roadway. 

(Ans. S 47° 19' W; S 55°19' E.) 

6. The following are the particulars of 3 boreholes. 



Borehole 


Surface Level A.O.D. 


Depth of Borehole 


A 


600 ft 


500 ft 


B 


400 ft 


600 ft 


C 


1000 ft 


600 ft 



If the distance from A to B is 1200 ft and from B to C 1800 ft, 
calculate the gradiants of the lines AB and BC. 

(Ans. 1 in 4; 1 in 3) 

7. A and B are two boreholes which have been put down to prove a 
seam. They are on the line of full dip of the seam, the direction of 
line BA being N 50°E and its plan length 1000 ft. 



Borehole 



Surface Level 



Depth 



A 600 ft 750 ft 

B 800 ft 700 ft 

A shaft is to be sunk at a point C, the surface level of C being 
1000 ft, the length BC 800 ft, and the direction of BC due East. 
Calculate (a) the dip of the seam from B to C, 
(b) the depth of the shaft at C. 

(Ans. (a) 1 in 5- 23; (b) 1053 ft) 

8. A seam dips at 1 in 12-75, S 17° W and at 1 in 12*41, S 20°15' E 
Calculate the magnitude and direction of full dip. 

(Ans. 1 in 11-64; S 6°46' E). 

9. The co-ordinates and level values of points A, B and C respec- 
tively, in a mine, are as follows: 



Departure (ft) 



Latitude (ft) 



A +119-0 + 74-0 9872 

B -250-0 + 787-5 9703 

C +812-0 +1011-0 9805 

(a) Plot the positions of the points to a convenient scale and graphic- 
ally determine the direction and amount of full dip. 



Levels in ft above a datum 
10000 ft below O.D. 



DIP AND FAULT PROBLEMS 431 

(b) Calculate the direction and amount of full dip. 

(Ans. (a) N 38° W; 1 in 4-66. 

(b) N 37°56' W; 1 in 4-672) 

10. In a steep seam a roadway AB has an azimuth of 190° dipping at 
22° and a roadway AC has an azimuth of 351° rising at 16°. 

Calculate the direction and rate of full dip of the seam. 

(Ans. 225° 54' 50", 26° 31') 

11. A cross-measures drift, driven due South and dipping at 16°, 
passes through a bed of shale dipping due North at 27°. The distance 
between the roof and floor of the bed of shale measured on the floor of 
the drift is 56 ft. 

Calculate the true and vertical thickness of the bed. 

(Ans. 38-19 ft; 42-86 ft) 

12. A small colliery leasehold was proved by 3 bores. The surface 
level at each bore and the depth to the seam were as follows: 



Bore 


Surface Level 


Depth 


A 


30 ft above O.D. 


190 ft 


B 


20 ft above O.D. 


220 ft 


C 


10 ft above O.D. 


240 ft 



Bore B is 560 yds from A N 30° E and bore C is 420 yd from B 
S 60° E. Find graphically, or otherwise, the direction of strike and the 
rate of full dip in inches per yard. 

If a shaft is sunk 800 yd from A in a direction S 45° E at what 
depth below datum will it reach the seam ? 

(Ans. N 15° W; 1% in. per yd; 200-4 ft)- 

13. An underground roadway driven on the strike of the seam has a 
bearing S 30° E. The seam has a full dip of 8 in to the yd in a 
northerly direction. At a point A on the roadway another road is to be 
set off rising at 1 in 5*8. 

Calculate the alternative bearings on which this second road may 
be set off. 

(Ans. N 80° 54' W; S 20° 54' E) 

14. Three bores A, B and C have been put down to a coal seam. B 
is due north from A, 1000 ft, and C is N 76° W, 850 ft from A. The 
surface levels of the boreholes are the same. The depth of A is 700ft, 
of B 1250 ft and of C 950 ft. 

Calculate the direction and rate of full dip and the slope area in the 
triangle formed by the boreholes. 

(Ans. N 16° 49' W; 1 in 1-74; 475610 sq ft) 



432 SURVEYING PROBLEMS AND SOLUTIONS 

8.4 The Rate of Approach Method for Convergent Lines 

In Fig. 8.21, let AC rise from A at 1 in K. 
AD dip from A at 1 in M. 
AB be the horizontal line through A. 




Fig. 8-21 
AF = 1 unit 



then GF = y units 



and FE = -T-. units 

M 



Comparing similar triangles ADC and AEG, 

CD GE ± 1_ 
BA FA KM 



■'■ CD = BA {\ + h) 

and BA = 



CD 



I 1 
K + M 

Thus if 2 convergent lines CA and DA are CD vertically apart, 
the horizontal distance BA when they meet _ CD 

~ I I ( 8 ' 13 > 



W/zen tfie fines both dip from A (Fig. 8.22) 

B F A 







it i 




c 


M 


"ril 


S^ 




M-K\ 








\ \/ 








^^e 











Fig. 8 



DIP AND FAULT PROBLEMS 



433 



Let AC dip from A at 1 in K 
AD dip from A at 1 in M. 

As before, 



CD 
BA 



G£ 
FA 

CD 



and BA = 



_1_ _ 1 

M K 

1 

BA 

CD 
£_ 1 



\M K) 



Therefore, if the two convergent lines CA and DA are CD verti- 
cally apart, the horizontal distance BA when they meet 

CD 



1 1 

M~ K 



(8.14) 



Thus if a rise is considered +ve and dip -ve, then the general 
expression applies. 

Horizontal distance = Yggical distance apart 

1_^ j_ 
M " K 



(8.15) 



Example 8.13 Two seams of coal, 100 ft vertically apart, dip at 
1 in 6. Find the length of a (drift) roadway driven between the seams 
(a) at a rise of 1 in 4 from the lower to the upper and (b) at a fall of 
1 in 2 from upper to lower. 



JlPPer 
searn 


. > 

ft 


L i ■ 


s eorrj 


T 


100' ^».--"~~~~ " 

" — — — — Li2_e "^-^ 




c, 


c 



(a) In triangle ADB, 



Fig. 8.23 

AD falls at 1 in 6. 
BD rises at 1 in 4. 
AB = 100 ft 



434 



SURVEYING PROBLEMS AND SOLUTIONS 



Then, by Eq. (8.15), 



D,D = T 



AB 



-(4) 



i.e. plan length of drift 
inclined length of drift 



100 





1 1 

4 + 6 




12 x 100 




3+2 


= 


240 ft 




240 x V4 2 + l 2 




4 




240 x yl7 




4 


= 


247*2 ft 




(b) Similarly, by Eq.(8.15), 
AB 



C. C 



plan length 



100 



6 \ 2) 2 6 



C^C = ^ = 300 ft 



inclined length AC = 3 °° X ^ 5 = 336'0 ft 

Example 8.14 Two parallel levels, 200 ft apart, run due East and 
West in a seam which dips due North at 3 in. to the yard. At a point 
A in the lower level a cross-measures drift rising at 6 in. to the yard 
and bearing N 30° E is driven to intersect another seam, situated 200 
ft vertically above the seam first mentioned, at a point C. From a point 



DIP AND FAULT PROBLEMS 



435 



B in the upper level due South from A another cross-measures drift 
rising at 12 in. to the yard and bearing N 30° E is also driven to inter- 
sect the upper seam at a point D. 

Calculate the length and azimuth of CD. (M.Q.B.) 




(418-70, 725-20) 



21 227-13) 



(0,-200) 



Fig. 8.25 
Apparent dip of seam in direction of drift (Fig, 8.25) 

ByEq(84 > .,._,„ cot full dip 









cosine angled between 






= 


X * = 13-856. 
cos 30° 


To find length of drift AC, Fig. 8.26(a), 


Plan length 


AC 


200 


200 x 83-136 


1 1 


19-856 




6 ' 13-856 










= 837-39 ft 



436 



SURVEYING PROBLEMS AND SOLUTIONS 



To find length of drift BD, Fig. 8.26(b), 
Plan length BD = 



200 


200 x 41-568 


1 1 


16-856 


3 + 13-856 


= 493-21 ft 



(a) 



200' 



UP 13-856 




837-39 



A '^-?--'" 




Fig. 8.26 

Assuming the two levels are 200 ft apart in plan, the relative 
positions of A, B, C and D can now be found from the co-ordinates 
(see chapter 3). 

AC N 30° E 837-39 ft 

.-. P.D. = +418-70 
.'. P.L. = +725-20 
Total departure of C = + 418-70 
Total latitude of C = +725-20 



sin 30° = 0-50000 
cos 30° = 0-86603 
(relative to A) 



AB due South 200 ft 



BD N 30° E 493-21 ft 
sin 30° = 0-50000 
cos 30°= 0-86603 



Total departure of B = 0*0 

Total latitude of B = -200-0 

.-. P.D. = +246-21 
.-. P.L. = +427-13 
Total departure of D = +246-21 



DIP AND FAULT PROBLEMS 437 

Total latitude of D = +427-13 - 200 

= +227-13 

« • n- rn ♦ -i 246-21 - 418-70 
Beanng of line CD = tan 22y . 13 _ ^.^ 

-i -172-49 
" tan -498-07 
= tan -1 0-346 32 
= S 19° 06' 10" W 

498-07 498-07 

Length of line CD = cosl9 o 06 / l0 « = q-94493 

= 527-10 ft (horizontal length) 

If the inclined length is required, 

AC rises at 1 in 6 •'• the difference in level 

A - C = + ™^- = +139-56 ft 
6 

BD rises at 1 in 3 •*• the difference in level 

493-21 
B - D = +^\^ = +164-40 ft 

.-. level of D relative to B = +164*40. 

AB rises at 1 in 12 •*. the difference in level 

A-B = + ^ = +16-67 ft 

/. Level of D relative to C = +164-40 + 16*67 - 139-56 

= +41-51 ft. 
.-. Inclined length CD = ./(527-10 2 + 41-51 2 ) 

= V( 277 834 + 1723) 

= 528-7 ft 



8.5 Fault Problems 

8.51 Definitions 

A geological fault is a fracture in the strata due to strains and 
stresses within the earth's crust, accompanied by dislocation of strata. 
The direction of movement decides the nature of the fault. 

With simple displacement either the corresponding strata are forced 
apart giving a normal fault or movement in the opposite direction causes 
an overlap known as a reverse fault. Many variations are possible with 
these basic forms. 



438 



SURVEYING PROBLEMS AND SOLUTIONS 





MAM////////. 



Normal fault 



Reverse fault 



Fig. 8.27 

Fig. 8.27 only illustrates the end view and no indication is given 
of movement in any other direction. 

The following terms are used (see Fig. 8.27): 
FF % is known as the fault plane, 
B is the upthrow side of the fault, 
C is the downthrow side of the fault, 
6 is the angle of hade of the fault (measured from the ver- 
tical), 
BE is the vertical displacement or throw of the fault, 
EC is the horizontal displacement, lateral shift or heave, 
causing an area of want or barren ground in the normal 
fault. 
Faults which strike parallel to the strike of the bed are known as 
strike faults. 

Faults which strike parallel to the dip of the beds are known as 
dip faults. 

Faults which strike parallel to neither dip nor strike are known as 
oblique faults and are probably the most common form, frequently with 
rotation along the fault plane, Fig. 8.28 . 



ab — strike slip 

be = dip slip 

ac = net or resultant slip 

bd = vertical throw 



Fig. 8.28 Diagonal or oblique fault 




DIP AND FAULT PROBLEMS 



439 



Where the direction and amount of full dip remain the same on both 
sides of the fault, the fault is of the simple type and the lines of con- 
tact between seam and fault on both sides of the fault are parallel. 

Where the direction and/or the amount of dip changes, rotation of 
the strata has taken place and the lines of contact will converge and 
diverge. The vertical throw diminishes towards the convergence until 
there is a change in the direction of the throw which then increases, 
Fig. 8.29. 



Down throw 




Down throw 



Fig. 8.29 

N.B. The strike or level line of a fault is its true bearing, which 
will differ from the bearing of the line of contact between seam and 
fault plane. 

Example 8- 15 A vertical shaft, which is being sunk with an excavated 
diameter of 23 ft 6 in. passes through a well-defined fault of uniform 
direction and hade. 

Depths to the fault plane below a convenient horizontal plane are 
taken vertically at the extremities of two diameters AB and CD, which 
bear north-south and east-west, respectively. The undernoted depths 

were measured: 

at A (north point) 10' 1" 

at B (south point) 26' 3" 

at C (east point) 4' 0" 

at D (west point) 32' 4" 

Calculate the direction of the throw of the fault and the amount of 
hade. Express the latter to the nearest degree of inclination from the 
vertical. (M.Q.B./S) 



N (10-1") 



(-32'-4")W 



Fig. 8.30 




E (4'-0") 



S(-26 -3 ) 



440 



SURVEYING PROBLEMS AND SOLUTIONS 




Fig. 8.31 
Gradient NS line 

(26'3"-10'l") in 23' 6" 

i.e. 16'2"in23'6" = 1 in 1-453 57 

.-. j3 = 34°31'40" 

Gradient EW line 

(32'4"-4'-0") in 23'-6" 
i.e. 28'-4"in 23'6" * =1 in 0-82951 

a = 50° 19' 30" 

From Eq. (8.6), 

tan A = tan /3 cosec 6 cot a - cot 6 

but 6 = 90° a tan A = tan£ cot a (by Eq. 8.9) 

= tan 34°31'40" x 0-829 51 

= 0-570 70 

8 = 29°42'50" 
•*. Bearing of full dip 

= 270° - 29°42'50" 

= 24 0° 17' 10" 

From Eq.(8.7), 

cot 8 = cot a cos A 

= 0-829 51 cos29°42'50" 

= 0-720 44 

8 = 54° 13' 50" 

/. Angle of hade = 90 - 8 

= 35°46' 10" (36° to nearest degree) 



Example 8- 16 A roadway advancing due West in a level seam encoun- 
ters a normal fault running North and South, with a hade of 30°, which 



DIP AND FAULT PROBLEMS 



441 



throws the seam up by a vertical displacement of 25 ft. 

A drift rising at 1 in 3*6 is driven from the point where the road- 
way meets the fault on the East side to intersect the seam on the West 
side of the fault. 

Find, by drawing to scale, the inclined length of the drift and 
check your answer by calculation. 




Fig. 8.32 
Length of drift scaled from drawing = 93*5 ft. 

BC 25 



Length of drift 
but 



AB = 

sin (7 

cot(9 = 3*6 

= 15°31' 

25 



AB 



Length of drift 



sinl5°31 
= 93-4 ft. 



id 



= 93-42 ft. 



Example 8.17 A roadway dipping 1 in 8 in the direction of full dip of 
a seam strikes an upthrow fault, bearing at right-angles thereto. 
Following the fault plane a distance of 45 ft the seam is again located 
and the hade of the fault proved to be 30°. 

Calculate the length of a cross-measures drift to win the seam, 
commencing at the lower side of the fault and rising at 1 in 6 in the 
same direction as the roadway. 




Fig. 8.33 

Vertical throw of fault FB = 45 cos 30° = 38*97 ft 
Lateral displacement FC = 45 sin 30 = 22' 50 ft 



442 



SURVEYING PROBLEMS AND SOLUTIONS 



In triangle EFC 

EF = 



EC 



22-50 



= 2-81 ft 



8 8 

.-. EB = FB + EF = 38-97 + 2-81 = 41-78 ft. 
To find the plan length of the drift by the rate of approach method, 



BG = 



41-78 
1 1 
8 + 6 



= 143-24 ft 
Inclined length of drift BD 

= 143-24 x 



41-78 x 48 
14 



= 145-22 ft 



Example 8.18 A roadway, advancing due East in the direction of 
full dip of 1 in 8, meets a downthrow fault bearing S 35° E, with a 
throw of 60 ft and a hade of 30°. 

At a distance 150 ft along the roadway, back from the fault, a 
drift is to be driven in the same direction as the roadway in such a way 
that it meets the point of contact of seam and fault on the downthrow 
side. 

Calculate the inclined length and gradient of the drift, assuming 
the seam is constant on both sides of the fault. 



Section 

on fault 60' 




Fig. 8.34 



DIP AND FAULT PROBLEMS 443 

In plan 

Width of barren ground = BX = 60 tan 30° = 34-64 ft. 

On the line of the roadway BC = 34-64 sec 35° = 42- 29 ft (= QC). 

As X is 60 ft vertically below B, it is necessary to obtain the 
relative level of C. 

XC = BC sin 35° 

XY = XC sin 35 = BC sin 2 35° = 42- 29 sin 2 35° = 13-91 ft 

As the dip of XY is 1 in 8, the level of Y relative to X is 

O 

But YC is the line of strike 
.'. Level of Y = level of C = 1-74 ft below X 

= 60 + 1-74 ft below A 
= 61-74 ft below A 

(i.e. BQ = MP). 
In section 

Gradient of roadway AB = 1 in 8. 
.-. cot0 = 8 

= 7°08' 
Thus AM = 150sin7°08' = 18-63 ft 
and MB = 150 cos7°08' = 148-84 ft 

(= PC). 
Difference in level A - C = AP = AM + MP 

= 18-63 + 61-74 = 80-37 ft 

Plan length of drift AC = PC = PQ + QC 

= 148-84 + 42-29 = 191-13 ft 

••• Gradient of drift = 80-37 ft in 191-13 ft 

, . 191-13 

= luk ~WW 

= 1 in 2-378 

cot<£ = 2-378 

<f> = 22°49' 

Length of drift = 19 1- 13 sec 22°49 ' 

= 207-36 ft 

8.52 To find the relationship between the true and apparent bearings 
of a fault 

The true bearing of a fault is the bearing of its strike or level line. 
The apparent bearing of a fault is the bearing of the line of con- 



444 SURVEYING PROBLEMS AND SOLUTIONS 

tact between the seam and the fault. 
The following conditions may exist: 

(1) If the seam is level, the contact line is the true bearing of the 
fault. 

(2) The apparent bearings are alike, i.e. parallel, if the full dip of 
the seam is constant in direction and amount. N.B. The throw of the 
fault will also be constant throughout its length— this is unusual. 

(3) The apparent bearings differ, due to variation in direction or 
amount of full dip on either side of the fault. The barren ground will 
thus diminish in one direction. N.B. The throw of the fault will vary 
and ultimately reduce to zero and then change from upthrow to down- 
throw. This is generally the result of rotation of the beds. 

Two general cases will therefore be considered: 

(1) When the throw of the fault opposes the dip of the seam. 

(2) When the throw of the fault is in the same general direction as 
the dip of the seam. 

Let the full dip on the downthrow side be 1 in x, 
the full dip on the upthrow side be 1 in y, 
the angle between the full dip and the line of contact be a, 
the angle between the line of contact and the true bearing of 
the fault be /3, 

the angle of hade be 6° from the vertical, 
the throw of the fault be t ft down in the direction NW, 
the angle between the full dip 1 in y and the true bearing of 
the fault be <f>. 

8.53 To find the true bearing of a fault when the throw of the fault 
opposes the dip of the seam (Fig. 8.35) 

If the throw of the fault is t ft, then D will be + 1 ft above A, 
and for the true bearing of the fault DC, C must be at the same level 
as D. Angle ADC = 90° with DC tangential to the arc of radius t . tan Q. 
The full dip 1 in x requires a horizontal length AB = tx ft. 

The same applies on the other side of the fault. E must be at the 
same level as A, and EA, the true bearing of the fault, must be paral- 
lel to DC; also for FE to be the strike in the seam DF must equal 
ty ft. 

Referring to Fig. 8.35, 

In triangle ABC, 

AC = tX 



cos a 



DIP AND FAULT PROBLEMS 

In triangle ACD, 

t tan0 



445 



sinjS, = — = tan0 cos a, (8.16) 



tx 



cos a, 



In triangles ADE and EDF, 



but 



. tan0 coscl, 
smfl, = - 

y 

a 2 = <£ + j8 2 

. tan0 

•• sin P 2 - — — cos(<£+/8 2 ) 



(8.17) 



(8.18) 



tan0 



(cos tfi cos /3 2 - sin sin /3 2 ) 



i.e. y cot# = cos <f> cot /3 2 - sin <f> 
y cot + sin <f> 



cotj8 2 



COS0 



(8.19) 



Hence the true bearing of the fault 

= bearing of contact line AC - /3, 



Bearing of contact line ED 



(8.20) 

= true bearing of fault + /3 2 

= bearing of AC - ft, + ft 2 (8.21) 




Fig. 8.35 



446 



SURVEYING PROBLEMS AND SOLUTIONS 



8.54 Given the angle 8 between the full dip of the seam and the true 
bearing of the fault, to find the bearing of the line of contact 

(Fig. 8.35) 



From Eq.(8.16), 



sin 



a, = 8+P, 

tanO cos a, 
x 

tan0 /o. o \ 
= cos (5 + ft,) 



x cot# = 



cos g cos ft, - sin 8 sinft, 



cot ft, 



sinp^ 
= cos 5 cot ft, - sin 8 

x cot d + sin 8 
~ cos 8 



(8.22) 



Example 8.19 A plan of workings in a seam dipping at a gradient of 

S W 
1 in 3 in a direction S 30° E shows a fault bearing 45 F in the seam 

which throws the measures down to the North-West. The hade 

of the fault is 30° to the vertical. Calculate the true bearing of the 

fault. 




From Eq.(8.16), 

sin ft = 



Fig. 8.36 

cos a tanfl _ cos 75° tan 30° 
x 3 

0-149430 



= 0-049 810 
ft = 2°51'20" 
/. True bearing = 045°00'00" - 2°51'20" 



o r\Q> An" 



= 042° 08' 40 



DIP AND FAULT PROBLEMS 



447 



Example 8.20 A seam dipping 1 in 5, S 60° E, is intersected by a 
fault the hade of which is 30° to the vertical and the bearing N 35° W. 
The fault is a downthrow to the South- West. Calculate the bearing of 
the fault as exposed by the seam. 




Fig. 8.37 



As full dip and hade oppose each other, by Eq. (8.22) 

x cot 6 + sin 8 



cot/3 = 



= 



cos 5 

5 cot30 o + sin25° 

cos 25° 
10-02181 
5°42' 



Bearing of fault exposed in seam 

= 325°- 5° 42' 
= 319° 18' = N40°42 / W 
Example 8.21 Headings in a seam at A and B have made contact 
with a previously unlocated fault which throws the measures up 100 ft 
to the south-east with a true hade of 40° from the vertical. 

Full dip is known to be constant in direction, namely 202° 30' , 
but the amount of dip changes from 1 in 5 on the north side to 1 in 3 
on the south side of the fault. 

Given the co-ordinates of A and B as follows: 
A 3672-46 ft E. 5873-59 ft N. 
B 4965-24 ft E. 7274*38 ft N. 
Calculate (a) the true bearing of the fault, (b) the bearings of 
the lines of contact between fault and seam. 



448 



SURVEYING PROBLEMS AND SOLUTIONS 

/a 



% / 




4oV 



Section 
on fault 



Fig. 8.38 

Co-ordinates E N 

A 3672-46 5873*59 

B 4965-24 7274-38 

dE + 1292-78 dN + 1400-79 

tan bearing AB = f = ±f|§f = 0-92289 

bearing of contact line AB = 042° 42' 

Then a, = 180° + 042°42' - 202°30' = 20°12' 

By Eq.(8.16), 

. Q cos20°12' tan 40° 
sm/3, = = = 0-157 50 

j8, = 9°04' 
Bearing of line of strike of fault, 
i.e. true bearing of fault = 042°42' - 9°04' 

= 033°38' 

Now <fi = a,-/3, = 20 o 12'-9 o 04 , = 11°08' 
By Eq.(8.19) 



cotjS 2 = 



3 cot40° + sinll o 08' 



cosll°08' 
= 3-840 62 
j8 2 = 14°34' 
Bearing of line of contact on upthrow side of fault 
= 033°38' + 14°34' = 048°12' 



DIP AND FAULT PROBLEMS 



449 



8.55 To find the true bearing of a fault when the downthrow of the 
fault is in the same general direction as the dip of the seam 

(Fig. 8.39) 




Fig. 8.39 

. a tan 6 cos a, 
smjS, = - 



before 



and sin j6 2 = 



tan# cosa.j 



(8.23) 
(8.24) 



but 



a * = <f>~Pz 



sin)8 2 = 



tan0 



(cos cf> cos 2 + sin cf> sin 2 ) 



y cotd = cos<£ cot/S 2 + sin<£ 

. a y cot - sin cb 

cotj8 2 = ^ 



(8.25) 
(8.26) 



COS0 

Hence true bearing of fault = bearing of contact line CA + j8, 
bearing of contact line DE = true bearing of fault - j8 2 

= bearing of CA + ff, - /3 2 (8.27) 



8.56 Given the angle 5 between the full dip of the seam and the true 
bearing of the fault, to find the bearing of the line of contact 



Here 
i.e. 



8 = a 1+ j8, 
a, = 5,-0, 



450 SURVEYING PROBLEMS AND SOLUTIONS 

then, from Eq.(8.23), 

sin/3, = -j- cos(5-£,) 

L „ x cot0 - sinS, 
giving cot/3, = ^ 



(8.28) 



Example 8-22 A fault is known to be an upthrow in a NW direction 
with a full hade of 30° to the vertical. 

The bearing of the fault as exposed in the seam is N 40° E, and 
the full dip of the seam is 1 in 5 S 35° W. Find the true bearing of the 
fault. 




Fig. 8.40 

As the downthrow of the fault is in general direction as the full 

dip, by Eq. (8.23), 

. „ tan0 cos a 
sin0 = - 

tan 30° cos 5° 



5 
= 0-11503 

B = 6°36' 
.-. Bearing of fault = N 40° + 6°36' E 
= N 46 36 y E 

8.6 To Find the Bearing and Inclination of the line 
of Intersection (AB) of Two Inclined Planes 

Let (1) the horizontal angle between the lines of full dip of the 
two planes inclined at a and j8 respectively be S, 
(2) the horizontal angle between the line of full dip a and 
the intersection line AB of inclination <f> be 0. 



DIP AND FAULT PROBLEMS 



451 




Fig. 8.41 

From Eq. (8.5), 

tan <f> = tan a cos 6 

also = tanjS cos(S-0) 

tan a cos = tan B (cos 8 cos 6 + sin 5 sin 0) 

i.e. tan a cot j8 = cos 8 + sin 5 tan 6 

„ tan a cot B - cos 8 

tan0 = t 1 -^ — 

sin 3 



(8.29) 



If plane (a) is a plane in the form of a seam and plane (b) is a 
fault plane of hade (90 - B) 



it a tan a tan B - cos 8 

then tan0 = £-=: 

sino 



(8.30) 



Example 8.23 A is a point on the line of intersection of two inclined 
planes. The full dip of one plane is 1 in 8 in a direction 222° 15' and 
the full dip of the other is 1 in 4 in a direction 145°25' . 

If B is a point to the south of A on the line of intersection of the 
two planes, calculate the bearing and the inclination of the line AB. 

Let the angle of the full dip of plane (1) be a = cot _1 4 = 14°02' 

of plane (2) be B. 
Then the horizontal angle 8 between them = 222° 15' - 145° 25' 

= 76°50' 



From Eq.(8.29), 



a tana cot B - cos 8 

tan0 = f— =- 

sin 5 



i.e. tan0 = 



8 tanl4°02'-cos76°50' 
sin76°50' 

1-99960 -0-22778 



0-97371 
= 1-81966 



452 SURVEYING PROBLEMS AND SOLUTIONS 

6 = 61°12'30" 
.'. Bearing of AB = 145°25'+ 61°12'30" 

= 206°37 , 30" 
Angle of inclination AB(<f)) 

tan <f> = tan a cos 6 
= 0-120 38 
cf> = 6°52' 
i.e. 1 in 8*3 

Exercises 8(b) (Faults) 

15. A heading in a certain seam advancing due East and rising at 

6 in. to the yd met a fault with a displacement up to the East and the 
seam has been recovered by a cross-measures drift rising at 18 in. to 
the yd in the same direction as the heading. The floor of the seam 
beyond the fault where the gradient is unchanged was met by the roof 
of the drift when it had advanced 345 ft, and the roof of the seam when 
it had advanced 365 ft 

Calculate the thickness of the coal seam. 

(M.Q.B./M Ans. 5 ft 10^ in.) 

16. Two parallel seams 60 ft vertically apart dip due W at 1 in 6. 
A drift with a falling gradient of 1 in 12 is driven from the upper to 
the lower seam in a direction due E. 

Calculate the length of the drift. 

(Ans. Hor. 240 ft; Incl. 240-82 ft). 

17. A roadway in a level seam advancing due N meets a normal 
fault with a hade of 30° from the vertical and bearing at right-angles 
to the roadway. 

An exploring drift is set off due N and rising 1 in 1. At a dis- 
tance of 41 ft, as measured on the slope of the drift, the seam on the 
north side of the fault is again intersected. 

(a) Calculate the throw of the fault and the width of the barren ground. 

(b) If the drift had been driven at 1 in 4 (in place of 1 in 1) what 
would be the throw of the fault and the width of barren ground ? 

(Ans. (a) 29 ft; 16-7 ft 
(b) 9-9 ft; 5-7 ft) 

18. A roadway AB, driven on the full rise of a seam at a gradient of 
1 in 10, is intersected at B by an upthrow fault, the bearing of which 
is parallel with the direction of the level course of the seam, with a 
hade of 30° from the vertical. 

From B a cross-measures drift has been driven in the line of AB 
produced, to intersect the seam at a point 190 ft above the level of B 



DIP AND FAULT PROBLEMS 453 

and 386 ft from the upper side of the fault as measured in the seam. 

Calculate the amount of vertical displacement of the fault and the 
length and gradient of the cross-measures drift. Assume that the direc- 
tion and rate of dip of the seam is the same on each side of the fault. 

(Ans. 151-6 ft; 507-78 ft; 1 in 2-5) 

19. A roadway advancing due East in a level seam meets a fault bear- 
ing North and South, which hades at 30°. A drift, driven up the fault 
plane in the same direction as the roadway, meets the seam again at a 
distance of 120 ft. 

Calculate the length and gradient of a drift rising from a point on 
the road 400 ft to the West of the fault which intersects the seam 100 
ft East of the fault. 

(Ans. 570 ft; 1 in 5-4) 

20. The direction and rate of full dip of two seams 60 yd vertically 
apart from floor to floor are N 12° E and 4 l / 2 in. to the yd respectively. 

Calculate the length of a cross-measures drift driven from the 
lower seam to intersect the upper, and bearing N 18° W. (a) if the 
drift is level; (b) if it rises at a gradient of 4 in. to the yd. 

(M.Q.B./M Ans. (a) 1662-7 ft; 
(b) 825-2 ft) 

21. A level roadway AB bearing due West, in a seam 8 ft thick, 
strikes a normal fault at B, the point B being where the fault plane 
cuts off the seam at floor level. 

The hade of the fault, as measured in the roadway is 35° from the 
vertical. A proving drift is driven on the same bearing as the roadway 
and dipping from the point B at 1 in 3. The floor of the drift intersects 
the floor of the seam, on the lower side of the fault at a point C and 
BC is 80 ft measured on the slope. At C the seam is found to be 
rising at 1 in 10 due East towards the fault. 

Draw a section to a scale of 1 in. = 20 ft on the line ABC show- 
ing the seam on both sides of the fault, the drift BC and the fault plane 
and mark on your drawing the throw of the fault and the distance in the 
seam from C to the fault plane. 

(M.Q.B./M Ans. 19 ft; 63 ft) 

22. The direction of full dip is due North with a gradient of 1 in 6 
on the upthrow side and 1 in 9 on the downthrow side. Workings show 
the line of contact of the seam and fault on the upthrow side as 

N 45° E with a fault hade of 30° and throw of 30 ft. 

Calculate (a) the true bearing of the fault, (b) the bearing of the 
line of contact on the downthrow side. 

(Ans. 041°06'; 043°45') 



454 SURVEYING PROBLEMS AND SOLUTIONS 

Exercises 8(c) (General) 

23. In a seam a roadway AB on a bearing 024°00' dipping at 1 in 9 
meets a second roadway AC bearing 323°00' dipping at 1 in 5 x /2. 

Calculate the rate and direction of full dip of the seam. 

(Ans. 331°13'; 1 in 5-44) 

24. Two seams 60 ft vertically apart, dip at 1 in 8 due N. It is re- 
quired to drive a cross-measures drift in a direction N 30° W and 
rising at 1 in 5 from the lower to the upper seam. 

Calculate the length of the drivage. 

(Ans. 198-5 ft) 

25. A roadway driven to the full dip of 1 in 12 in a coal seam meets 
a 240 ft downthrow fault. A cross-measures drift is set off in the di- 
rection of the roadway at a gradient of 6 in. to the yd. 

In what distance will it strike the seam again on the downthrow 
side if (a) the hade is 8° from the vertical, (b) the hade is vertical ? 

(Ans. (a) 2885-7 ft; (b) 2919-8 ft) 

26. Three boreholes A, B and C intersect a seam at depths of 
540 ft, 624 ft and 990 ft respectively. A is 1800 ft North of C and 
2400 East of B. 

Calculate the rate and direction of full dip. 

(Ans. 1 in 3-96; S 7°58' W) 

27. The rate of full dip of a seam is 4^ in. to the yd and the direc- 
tion is S 45° E. 

Find, by calculation or by drawing to a suitable scale, the inclina- 
tions of two roadways driven in the seam of which the azimuths are 
195° and 345° respectively. 

(Ans. 1 in 16; 1 in 9-24) 

28. Two parallel roadways AB and CD advancing due North on the 
strike of a seam are connected by a road BD in the seam, on a bearing 
N 60° E. 

The plan length of BD is 150 yd and the rate of dip of the seam 
is 1 in 5 in the direction BD. 

Another roadway is to be driven on the bearing N 45° E to connect 
the two roadways commencing at a point 200 yd out by B on the road 
AB, the first 20 yd to be level. 

Calculate the total length of the new roadway and the gradient of 
the inclined portion. 

(Ans. 186-44 yd; 1 in 5*46) 

29. A roadway AB, 700 ft in length, has been driven in a seam of 
coal on an azimuth of 173°54'. It is required to drive a cross-measures 
drift from a point C in another seam at a uniform gradient to intersect 
at a point D, the road AB produced in direction and gradient. The 



DIP AND FAULT PROBLEMS 455 

levels of A, B and C in feet below Ordnance Datum and the co-ordin- 
ates of A and C in feet are respectively as follows: 
Levels Latitude Departure 
A -1378 +9209 +18041 
B -1360 

C -1307 +6180 +17513 
Calculate the length AD, and the length and gradient of the propo- 
sed drift CD, assuming that the latter is to have an azimuth of 032°27'. 

(Ans. 1892-9 ft; 1359-0 ft; 1 in 3-27) 

30. A heading AB driven direct to the rise in a certain seam at a 
gradient of 6 in. to the yd and in the direction due N is intersected 
at B by an upthrow fault, bearing at right-angles thereto with a hade 
of 30° from the vertical. From the point B a cross-measures drift has 
been driven in the direction of AB produced, and intersects the seam 
at a point 420 ft from the upper side of the fault. The levels at the 
beginning and end of the cross-measures drift are 100 ft and 365 ft 
respectively above datum. 

Calculate (a) the vertical displacement of the fault, (b) the 
length and gradient of the drift. 

Assume the direction and rate of dip of the seam to be uniform on 
each side of the fault. 

(M.Q.B./M Ans. (a) 195-9 ft (b) 590-2; 1 in 2) 

31 . A roadway in a seam dipping 1 in 7 on the line of the roadway 
meets a downthrow fault of 30 ft with a hade of 2 vertical to 1 horizon- 
tal. 

Calculate the length of the drift, dipping at 1 in 4 in the line of 
the roadway, to win the seam, a plan distance of 50 ft from the dip 
side of the fault; also the plan distance from the rise side of the fault 
from which the drift must be set off. 

Assume the gradient of the seam to be uniform and the line of the 
fault at right- angles to the roadway. 

(Ans. 267-97 ft; 194-97 ft) 

32. A roadway, bearing due East in a seam which dips due South at 

1 in 11, has struck a fault at a point A. The fault which, on this side, 
runs in the seam at N 10° W is found to hade at 20° from the vertical 
and to throw the seam down 30 ft at the point A. The dip of the seam 
on the lower side of the fault is in the same direction as the upper side 
but the dip is 1 in 6. 

From a point B in the roadway 140 yd West of A a slant road is 
driven in the seam on a bearing N 50° E and is continued in the same 
direction and at the same gradient until the seam on the East side of 
the fault is intersected at a point C. 



456 



SURVEYING PROBLEMS AND SOLUTIONS 



Draw to a scale of 1 in. to 100 ft a plan of the roads and fault and 
mark the point C. State the length of the slant road BC. 

(M.Q.B./S Ans. 264 yd) 
33. The sketch shows a seam of coal which has been subjected to 
displacement by a trough fault. 

Calculate the length and gradient of a cross-measures drift to con- 
nect the seam between A and B from the details shown. 

(Ans. 587-2 ft; 1 in 2-74) 

X Y 700' 




Ex. 8.33 

34. A seam dips 1 in 4, 208°30'. Headings at A and B have proved 
the bearing of the contact line AB to be 075°00*. 

If the hade of the fault is 30°, what is the true bearing of the fault 
if (a) it is a downthrow to the South; (b) it is a downthrow to the 
North. (Ans. 080°42'; 069°18') 

35. (a) Define the true and apparent azimuth of a fault. 

(b) A fault exposed in a certain seam has an azimuth of 086°10', 
and a hade of 33°. It throws down to the North West. The full dip of 
the seam is 1 in 6- 5 at 236° 15'. 

Calculate the true azimuth of the fault. Check by plotting, 
(c) Two seams, separated by 86 yd of strata, dip at 1 in 13 in a 
direction S 36° W. They are to be connected by a drift falling at 1 in 5, 
N 74°30'E. 

Calculate the plan and slope length of the drift. 

(N.R.C.T. Ans. (b) 081°12' (c) 330-51 yd 337-07 yd) 

36. A mine plan shows an area of 3*6 acres in the form of a square. 

Measured on the line of full dip underground the length is 632*4 links. 

Calculate the rate of full dip. ,. - . „ ..„o„wv 

F (Ans. 1 in 3 or 18°24 ) 



Bibliography 

M.H. HADDOCK, The Location of Mineral Fields (Crosby Lockwood). 
R. McADAM, Colliery Surveying (Oliver & Boyd). 

FORREST, S.N. Mining Mathematics (Senior Course) (Edward Arnold). 
METCALFE, J.E., A Mining Engineer's Survey Manual (Electrical 

Press, London). 
M.H. HADDOCK, The Basis of Mine Surveying (Chapman & Hall). 



9 AREAS 

9.1 Areas of Regular Figures 

The following is a summary of the most important formulae. 

9.11 Areas bounded by straight lines 

Triangle (Fig. 9.1) 




Fig. 9.1 Triangle 

(a) (Area) A = half the base x the perpendicular height 

i.e. A = 1/2 Mi (9.1) 

(b) A = half the product of any two sides x the sine of 

the included angle 
i.e. A = 1/2 ab sin C = 1/2 ac sin B = 1/2 be sin A (9.2) 

(c) A = yfs(s-a)(s-b)(s-c) (9.3) 

where s= Wz(a + b+c). 

Quadrilateral 

(a) Square, A = side 2 or 1/2 (diagonal 2 ) (9.4) 

(b) Rectangle, A = length x breadth (9.5) 

(c) Parallelogram (Fig. 9.2), (i) A = axh (9.6) 

(ii) A = absina (9-7) 




/ b \ 


\' 


f . 



Fig. 9.3 Trapezium 



Fig. 9.2 Parallelogram 
(d) Trapezium (Fig. 9.3) 

A = half the sum of the parallel sides x the per- 
pendicular height 



458 



SURVEYING PROBLEMS AND SOLUTIONS 



i.e. A = 1/z (a + b)h 
(e) Irregular quadrilateral (Fig. 9.4) 

(i) The figure is subdivided into 2 triangles, 
A = 1/2 {AC x Bb) + uz(AC x Dd) 
= VzAC(Bb + Dd) 

a c B 



(9.8) 



(9.9) 





Fig. 9.4 Irregular quadrilateral 



Fig. 9.5 



(ii) A = i/ 2 (/lCxBD)sin0 
This formula is obtained as follows, Fig. 9.5: 

A = uzAX.BX sind + uzBX.CX sin(180 - 0) 
+ 1/2 CX.DX sin 6 + 1/2 DX.AX sin (180 - 0) 
As sin 6 = sin (180-0), 

A = 1/2 sin 6 [AX(BX + DX) + CX(BX + DX)] 

= 1/2 sin 6 [ (AX + CX) (BX + DX)] 
A = i/zsind AC xBD 



Regular polygon (Fig. 9.6) 

(i) A = s/zarxn (9.11) 



(9.10) 



/J 

(ii) As a = 2r tan-^- 



Area A = i/2rnx2rtan — 



= nr^tan — 
2 

2 360 „ «„ 
= nr 2 tan^— (9.12) 
2n 

(iii) A = 1/2 fl 2 sin 6 x n 

n 2 360 

= 2 R Sin ~n~ (913) 




Fig. 9.6 



Regular polygon 
(n sides) 



AREAS 

9.12 Areas involving circular corves 

Circle 

(i) A = nr z 

(ii) A = i/4 nd 2 
where it ~ 3-1416, 1/477 ~ 0-7854. 

Sector of a circle. (Fig. 9.7) 



459 



(9.14) 
(9.15) 




(i) A = rrr 2 

w 360 

(ii) A = ±f* 



Fig. 9.7 Sector of a circle 



e 



(9.16) 
(9.17) 



N.B. (ii) is generally the better formula to use as the radian value 
of is readily available in maths, tables or may be derived from first 
principles (see Section 2.22). 



Segment of a circle (Fig. 9.8) 




Fig. 9.8 Segment of a circle 



460 



SURVEYING PROBLEMS AND SOLUTIONS 



(i) A = area of sector — area of triangle 
i.e. A = i/2r 2 ^ d - i/2r 2 sin0 

= 1/2 r 2 (0 - sin 0) 



(9.18) 



or 



A = -nr z i/2r 2 sin(9 

360 



d 



= r "ll60- ,/2Sin ^ 



(9.19) 



(ii) If chord AC = c and height of arc = h are known, the 
approximation formulae are: 



(a) 
(b) 
or 



- Irfffl 



h 2 
2c 3 



A ^ 



W + 4c z h 
6c 



(9.20) 
(9.21) 
(9.22) 



Annulus (flat ring) (Fig. 9.9) 
A = 7rR 2 -7Tr z 

= n(R 2 -r 2 ) 

= 7r(/2-r)(/? + r) (9.23) 
or A = 1/4 Tr(D-d)(D+d) (9.24) 

where D = 2R, d = 2r. 



9.13 Areas involving non-circular curves 

Ei/ipse (Fig. 9.10) 




Fig. 9.9 Annulus 



77" 
A = -ab 
4 



(9.25) 




Fig. 9.10 Ellipse 



where a and ft are major and minor axes. 



AREAS 



461 



Parabola (Fig. 9.11) 



A = -bh 
3 



(9.26) 




Fig. 9.11 Parabola 



9.14 Surface areas 

Curved surface of a cylinder (Fig. 9.12) 





Fig. 9. 12 Curved surface area of a cylinder 

If the curved surface is laid out flat it will form a rectangle of 
length 2-77-r = rrD and of height h = height of cylinder. 



Surface area (S.A.) = lirrh 
S.A. = vDh 



(9.27) 
(9.28) 



Curved surface of a cone (Fig. 9.13) 




Fig. 9.13 Curved surface area of a cone 



462 



SURVEYING PROBLEMS AND SOLUTIONS 



If the curved surface is laid out flat it will form the sector of a 
circle of radius /, i.e. the slant height of the cone. 

By Eq. (9.17), 

Area of sector = 1/2 r 2 6 Tad = 1/2 rxr0 



= i/2r x arc 
which here = 1/2/ x ttD 
S.A. = 1/2 nlD 
i.e. 1/2 x circumference of the base of the cone x slant height. 



(9.29) 



Surface area of a sphere (Fig. 9. 14) 

This is equal to the surface area of a cylinder of diameter D = 
diameter of the sphere and also of height D. 




Fig. 9. 14 Surface areas of a sphere 
A = TTDxD 

= ttD 2 (9.30) 

or A = 277Tx2r = Airr 2 (9.31) 

Surface area of a segment of a sphere 

In Fig. 9.14, if parallel planes are drawn perpendicular to the axis 
of the cylinder, the surface area of the segment bounded by these 
planes will be equal to the surface area of the cylinder bounded by 
these planes. 

i.e. S.A. = nDh (9.32) 

S.A. = 2-rrrh (9.33) 

where h = the height of the segment. 



AREAS 



463 



N.B. The areas of similar figures are proportional to the squares of 



the corresponding sides 



(9.34) 



In Fig. 9. 15, 
Area A 4BC 



AB 2 

2 



Area A AB,C, AB\ 



BC'' 



B,C 2 



AC 
~ACf 



h 2 




A C, C 

Fig. 9. 15 Areas of similar figures 



Similarly, in Fig. 9.16, 
Area of circle 1 = irr 2 = i/4 7rd, 
Area of circle 2 = 7rr 2 2 = viirtg 

Area 1 



as before 



Area 2 



rl 
dl 



(9.35) 



(9.36) 




Fig. 9.16 Comparison of areas 
of circles 



Example 9.1. The three sides of a triangular field are 663*75 links, 
632*2 links and 654*05 links. Calculate its area in acres. 



a = 


663*75 


s -a = 311*25 


b = 


632*20 


s - b = 342*80 


c = 


654*05 


s-c = 320*95 



2U950-00 



= 975*00 



s = 



975*00 



464 SURVEYING PROBLEMS AND SOLUTIONS 



By Eq. (11.3), 



Area = V s(s - a)(s - b)(s - c) 



= V975-00 x 311-25 x 342-80 x 320-95 
= 182724 link 2 
= 1-827 acres 

Example 9.2 A parallelogram has sides 147-2 and 135-7 ft. If the 
acute angle between the sides is 34°32' calculate its area in square 
yards. 

By Eq. (11.7), 

Area = ab sin a 

= 147-2 x 135-7 sin 34°32' 

= 11323 ft 2 

= 1258 yd 2 

Example 9.3 The area of a trapezium is 900 ft . If the perpendicular 
distance between the two parallel sides is 38 ft find the length of the 
parallel sides if their difference is 5 ft. 

ByEq.(9.8), 3g 

A = \ x + (x - 5) { — = 900 

1800 

.-. 2x - 5 = — — 
38 

47-368 + 5 52-368 



x = 



2 2 

= 26-184 
Lengths of the parallel sides are 26-184 and 21-184. 

Example 9.4 ABC is a triangular plot of ground in which AB mea- 
sures 600 ft, (182-88 m). If angle C = 73° and angle B = 68° find the 
area in acres. 

In triangle ABC, 

AB sin A 600 sin {180 - (73 + 68)} 



BC = 



sin C sin 73 

600 sin 39° 



sin 73° 
= 394-85 ft (120-345 m) 



AREAS 



465 



ByEq.(9.2), 

Area of triangle 



^BCABsinB 

2 

-x 394-85x600 sin 68 

2 



109829 ft 2 

109829 
9 x 4840 
2*521 acres. 



(10 203 m 2 ) 



acres 



Example 9.5 Calculate the area of the underground roadway from the 
measurements given. Assume the arch to be circular. 



In the segment (Fig. 9. 18) 

h = 7'- 10"- 5' = 2'- 10" 
c = 12'- 0" 



"-tf*© 1 



= 


r 2 - 2rh + h z + -c z 




h 2 + ±c z 2-833 2 + ^ 

4 n 




2h 2 x 2-833 


= 


7-769 ft 


Sin- = 
2 


c 12 


2r 2 x 7-769 


= 


0-77 230 


e 

2 ~ 


50°33'40" 



6 = 101°07'20' 



/^ ^X % " 




o "Ks 

U) 

" " 


12-0" 




" 





Fig. 9. 17 





.» ,n 






^" 




^^^ ^ 






yr ^' 






^r ^^ 




/» N. 


Z''' 






X. c 






>v 2 


a 


r-/> >^ 


rX. 


2 





Fig. 9.18 



Area of segment = - r\6 - sin 0) 



7-769 2 



(1-76492-0-98122) 



= ■ 30-179 x 0-78370 

= 23 ' 651 ft2 
Area of rectangle = 5x12 = 60 ft 2 
Total area = 83-651 ft 2 



466 



SURVEYING PROBLEMS AND SOLUTIONS 



Check 

By Eq.(9.20), 



Area of segment 



ByEq.(9.22), 



Area of segment 



4 //5x2-833\ 2 . 
- fx 2-833 VV~I j +6 ' 

^ 3-776 V3-135 + 36 
~ 3-776 x 6-294 
~ 23-77 ft 2 



3 x 2-833 3 + 4 x 144 x 2-833 



6x12 



~ 23-6 lft 2 



Example 9.6 A square of 6 ft sides is to be subdivided into three 
equal parts by two straight lines parallel to the diagonal. Calculate 
the perpendicular distance between the parallel lines. 

Triangle ABC = - area of square 
Triangle BEF = - area of square 



Length of diagonal AC 


= 6yf2 


Length BJ 


= 3/2 




= 4-242 


Area of A BFE 2 




Area of A ABC 3 




BK 2 2 




BJ 2 3 





BK* = 



4-242' 
1-5 



BK = 



4-242 x y/VS 
VS 
= 3-464 ft 
KJ = BJ-BK 

= 4-242-3-464 
Width apart of parallel lines 




Fig. 9.19 



0-778 
1-556 ft 



AREAS 



467 



Example 9.7 The side of a square paddock is 65 yd (59*44 m). At 
a point in one side 19V2yd (17*83 m) from one corner a horse is tether- 
ed by a rope 39 yd (35*66 m) long. 

What area of grazing does the horse occupy? 



F B 




Fig. 9.20 

The area occupied consists of a right-angled triangle AEG + ^ 
sector EFG. 



19y 2 

In triangle AEG cos £ = — - =0*5 

£ = 60° 
ByEq.(9.2), 

Area of triangle AEG = \ x 19 x / 2 x 39 sin 60 

= 329*31 yd 2 (275*34 m 2 ) 
ByEq.(9.17), 

Area of sector 



1 x 39 2 x 120° 



'red 



Total area 



= 760*5x2*094 

= 1592*49 yd 2 (1331*52 m 2 ) 

= 1921*80 yd 2 (1606*86 m 2 ) of grazing 



Example 9.8 In order to reduce the amount of subsidence from the 
workings of a seam the amount of extraction is limited to 25% by driv- 
ing 12ft wide roadways. What must be the size of the square pillars 
left to fulfil this condition? 



468 



SURVEYING PROBLEMS AND SOLUTIONS 



JJI 



Pillar 



TF 



6'^- 



-fir 



ii c r 



Fig. 9.21 

N.B. The areas of similar figures are proportional to the squares of 
the relative dimensions, 



100 



i.e. 



(* + 12)' 



75 x 2 

i.e. 100x 2 = 75(x 2 + 24* + 144) 

= 75x 2 + 1800*+ 10800 
25x 2 - 1800* - 10800 = 
x 2 - 72x- 432 = 

By the formula for solving quadratic equations. 





-b±y/b 2 - 


4ac 




2a 




72 


, c = -432, a = 


1, 




72 +V5184 


+ 1728 


A." 


2 
72 ±V6912 
2 






= 36 ±41-57 


= 77-57 ft 



Alternative Solution 

Size of square representing 100% 
Size of square representing 75% 
Difference in size 
Actual difference in size 



Vioo = 10 


(Am 


V75 = 8-66 


(EF) 


= 1-34 




= 12 ft 





AREAS 



469 



If x is the size of the pillar, 1'34 : 12 

12 x 8-66 



8-66 : x 



1-34 
77-57 ft 



Example 9.9 A circular shaft is found to be 4 ft out of vertical at 
the bottom. If the diameter of the shaft is 20 ft, find the area of the 
crescent-shaped portion at the bottom of the shaft which is outside the 
circumference from the true centre at the surface. 




Fig. 9.22 



Area of crescent-shaped portion = Area of circle - 2 (area of seg- 
ment AXB). 



Area of circle 



= TTr 2 = 3-1416x100 = 314-16 ft 2 



Area of sector A XBQ = -r z 6 

1 2 



-i oy -i 2 

6/2 = cos — = cos — 

OA 10 



2 

e 

Area of sector AX BO, 

Area of triangle OAB 

Area of segment AXB 
Area of crescent 



78°32' 
157°04' 
1100x2-74133 



137-07 



Il00sinl57°04' = 19-48 

= 117-59 
314-16 - 2x117-59= 78-98 ft 2 



470 



SURVEYING PROBLEMS AND SOLUTIONS 



Example 9.10 In a quadrilateral ABCD.A = 55°10', B = 78°30' 
C = 136°20' . (A and C are diagonally opposite each other) AB = 
620 links, DC = 284 links. 

(a) Plot the figure to scale and from scaled values obtain the area 
in acres. 

(b) Calculate the area in acres. 



ISO-^' K» 78*30')«= 46*20' 




Construction (N.B. D = 360 - (55° 10' + 78°30' + 136°20') = 90°) 
Draw a line parallel to AD 284 links away. This will cut AE at C. 



From scaling, 



AC = 637 links 
Dd = 256 links 
Bb = 296 links 

552 

Area = \ (637 x 552) = 175810 sq links 
= 1*758 1 acres 



Calculation 

In triangle AEB, 



AE = 



AB sin B 620 sin78°30' 



sinfi 



sin46°20' 



= 839-89 Iks 



Area of triangle AEB = ^AE.AB sin A 

= 1839-89 x 620 sin 55°10' = 213713 sq Iks 



In triangle DEC, DE = DC cot 46°20' 
= 284cot46°20' 



= 271-08 Iks 



Area of triangle DEC = 



AREAS 
+ DE.DC 



1 271-08 x 284 

2 



Area of triangle ABCD = 213713 - 38493 



471 



= 38 493 sq Iks 

= 175 220 sq Iks 
= 1-7522 acres 



9.2 Areas of Irregular Figures 

In many cases an irregular figure can conveniently be divided into 
a series of regular geometrical figures, the total area being the sum of 
the separate parts. 

If the boundary of the figure is very irregular the following methods 
may be employed. 

9.21 Equalisation of the boundary to give straight lines (Fig. 9.24) 




Fig. 9.24 Equalisation of an irregular boundary 



472 SURVEYING PROBLEMS AND SOLUTIONS 

The irregular boundary A B C D E F is to be equalised by a line 
from a point on YY to F. 

Construction 

(1) Join A to C; draw a line Bb parallel to AC cutting YY at 
b. Triangle AbC is then equal to triangle ABC. 
(triangles standing on the same base and between the same 

parallels are equal in area). 

(2) Repeat this procedure. 

Join bD, the parallel through C to give c on YY. 

(3) This process is now repeated as shown until the final line eF 
equalises the boundary so that area eABX = area XFEDC 

N.B. (a) This process may be used to reduce a polygon to a triangle 
of equal area, 
(b) With practice there is little need to draw the construction 
lines but merely to record the position of the points b,c,d, 
e, etc. on line YY. 

Reduction of a polygon (Fig. 9.25) 




b A Ed 

Fig. 9.25 Reduction of a polygon to a triangle of equal area 

Area of triangle bCd = Area of polygon ABCDE. 

(4) Where the boundary strips are more tortuous the following 
methods may be adopted. 

9.22 The mean ordinate rule (Fig. 9.26) 

The figure is divided into a number of strips of equal width and 
the lengths of the ordinates o,, o 2 , o 3 etc. measured. 

(N.B. If the beginning or end of the figure is a point the ordinate is 
included as q = zero.) 



AREAS 



473 




Fig. 9.26 The mean ordinate rule 



The area is then calculated as 

(Ol+0 2 +0 3 + 4 + ...O n ) 

f\ = -^— — ^— — — -^^^— — — — — — — — - 

n 
where n = number of ordinates 

2 ordinates x W 



x(n-l)w 



or 



A = 



(9.37) 



(9.38) 



where W = 2 w. 

This method is not very accurate as it implies that the average 
ordinate is multiplied by the total width W. 

9.23 The mid-ordinate rule (Fig. 9.27) 

Here the figure is similarly divided 
into equal strips but these are then 
sub-divided, each strip having a 
mid-ordinate, aa, bb, cc etc. 

The average value of these mid- 
ordinates is then multiplied by the 
total width W. 





b . 


d 












a - 

^ j 










a^««. 




' S^i 


1 






6" 


C 

w 







i.e. Area = mW 

where m = the mean of the mean 
ordinates. (9.39) 

The only advantage of this meth- Fig- 9.27 The mid-ordinate rule 
od is that the number of scaled 
values is reduced. 



9.24 The trapezoidal rule (Fig. 9.28) 



01 


02 

. w » » 


03 

















Fig. 9.28 The trapezoidal rule 



474 



SURVEYING PROBLEMS AND SOLUTIONS 



This is a mote accurate method which assumes that the boundary 
between the extremities of the ordinates are straight lines. 



The area of the first trapezium A y 



«<m 



(o, + o 2 ) 



The area of the second trapezium A 2 



w, 



2 "<o 2 + o 3 ) 



The area of the last trapezium 4»-i = ~7T~ (°n-i + °n) 

If w, =w z = w 3 = w n = w; 

w 
then the total area = — [o, + 2o 2 + 2o 3 +...2o n _, + <^] 



(9.40) 



9.25 Simpsorfs rule (Fig. 9.29) 

This assumes that the boundaries 
are curved lines and are considered 
as portions of parabolic arcs of the 
form y = ax 2 + bx + c. 

The area of the figure Aab t cCB 
is made up of two parts, the trapez- 
ium AabcC + the curved portion 
above the line abc. 




Fig. 9.29 Simpsorfs rule 



The area of ab.cb = - of the parallelogram aa. b. c. cb. 

1 O 111 



4vv 



[<*> - {(°y + °3>] 



[1 -I 4\V r t I 

2(°i + ° 3 )J + ~ [°2 ~ 2<°i + °3)J 



W, 



= 3"t°i + 4o z + ° 3 1 



(9.41) 



N.B. This is of the same form as the prismoidal formula with the 
linear values of the ordinates replacing the cross-sectional areas. 



AREAS 475 

If the figure is divided into an even number of parts giving an odd 
number of ordinates, the total area of the figure is given as 

w 
A = [(o, + 4o 2 + o 3 ) + (03 + 4o 4 + 05)+... o n _ 2 + 4o n _, + qj 

w -■ 
.-. A = [q + 4o 2 + 2q, + 4o 4 + 2o 5 +... 2o n _ 2 + 4o n _, + qj (9.42) 

The rule therefore states that: 

a if the figure is divided into an even number of divisions, the tot- 
al area is equal to one third of the width between the ordinates mult- 
iplied by the sum of the first and last ordinate + twice the sum of the 
remaining odd ordinates + four times the sum of the even ordinates" 
This rule is more accurate than the others for most irregular areas and 
volumes met with in surveying. 

Example 9.11 A plot of land has two straight boundaries AB and 
BC and the third boundary is irregular. The dimensions in feet are 
AB « 720, BC = 650 and the straight line CA = 828. Offsets from CA 
on the side away from B are 0, 16, 25, 9, feet at chainages 0, 186, 
402, 652, and 828 respectively from A. 

(a) Describe briefly three methods of obtaining the area of such a 
plot. 

(b) Obtain, by any method, the area of the above plot in acres, ex- 
pressing the result to two places of decimals. 

(R.I.C.S.) 
The area of the figure can be found by 

(1) the use of a planimeter; 

(2) the equalisation of the irregular boundary to form a straight 
line and thus a triangle of equal area 

(3) the solution of the triangle ABC + the area of the irregular 
boundary above the line by one of the ordinate solutions. 




Fig. 9.30 



476 SURVEYING PROBLEMS AND SOLUTIONS 

By (2), Area = 1(754 x 625) = 235 625 ft 2 = 5-41 acres 



By (3), Area of triangle ABC = y/s(s - a)(s - b)(s - c) (Eq. 9.3) 
where s = -(a + b + c) 

i.e. a = 650 s-a = 449 

b = 828 s-b = 271 

c = 720 s-c = 379 



2)2198 s = 1099 Check 



s = 1099 



Area = V 109 ^ x 449 x 271 x 379 
= 225126 ft 2 



Area of the irregular boundary 

(a) By the mean ordinate rule, Eq. (9.38), 

+ 16 + 25 + 9 + 
Area = x 828 

= 8280 ft 2 
Total Area 233406 ft 2 



= 5*35 acres 

(b) By the trapezoidal rule, Eq. (9.40), 

Area = 1[186(0 + 16) + (402 - 186) (16 + 25) 

+ (652 - 402) (25 + 9) + (828 - 652) (9 + 0)] 
= 1[186 x 16 + 216 x 41 + 250 x 34 + 176 x 9] 

= \ [2976 + 8856 + 8500 + 1584] 
= 10958 ft 2 



Total Area 236 084 ft 2 
= 5*43 acres 

N.B. As the distance apart of the offsets is irregular neither the 
full Trapezoidal Rule nor Simpson's Rule are applicable. 



AREAS 



477 



9.26 The planimeter 

This is a mechanical integrator used for measuring the area of ir- 
regular figures. 

It consists essentially of two bars OA and AB, with fixed as 
a fulcrum and A forming a freely moving joint between the bars. Thus 
A is allowed to rotate along the circumference of a circle of radius 
OA whilst B can move in any direction with a limiting circle OB. 

Theory of the Planimeter (Fig. 9.31) 

Let the joint at A move to A y 
and the tracing point B move first 
to B, and then through a small 
angle 5a to B 2 . 

If the whole motion is very small, 
the area traced out by the tracing 
bar AB is ABB.B^ 



i.e. ABx 8h + ±AB z 8a 



or 



8A = I8h + jl 2 8a 



(9.43) 




where 8A = the small increment of 
area, 

/ = the length of the tracing 
bar AB, 

8h = the perpendicular height 
between the parallel 
lines AB and j4,B„ 

5a = the small angle of rot- 
ation. Fig. 9.31 Theory of the planimeter 

A small wheel is now introduced at W on the tracing bar which 
will rotate, when moved at right-angles to the bar AB, and slide when 
moved in a direction parallel to its axis, i.e. the bar AB. 

Let the length AW = k.AB. 
Then the recorded value on the wheel will be 

8W = 8h + kAB8a 

i-e. 8h = 8w _ kAB8a = 8w - kl8a 

which when substituted in Eq. (9.43) gives 

54 = l(8w-U8a) + U 2 8a 



= I8w+ Z 2 (i -k)8a 



(9.44) 



478 



SURVEYING PROBLEMS AND SOLUTIONS 



To obtain the total area with respect to the recorded value on the 
wheel and the total rotation of the arm, by integrating, 



A = lw + 



f(i-kja 



(9.45) 



where 



A = the area traced by the bar, 
w = the total displacement recorded on the wheel, 
a = the total angle of rotation of the bar. 
Two cases are now considered: 

(1) When the fulcrum (0) is outside the figure being traced. 

(2) When the fulcrum (0) is inside the figure being traced. 

(1) When the fulcrum is outside the figure (Fig. 9.32) 

Commencing at a the joint is at 
A. 

Moving to the right, the line abed 
is traced by the pointer whilst the 
bar traces out the positive area 
04,) abcdDCBA. 

Moving to the left, the line defa 
is traced out by the pointer whilst 
the negative area (4 2 ) defaAFED 
is traced out by the bar. 

The difference between these 
two areas is the area of the figure 
abedefa, i.e. 

Fig. 9.32 Planimeter fulcrum outside 
the figure 




A = 4, - A z = lw + i 



! (i-*)a 



but 



a = 

A = lw 



(Eq. 9.45) 



(9.46) 



N.B. The joint has moved along the arc A BCD to the right, then 
along DEFA to the left. 

In measuring such an area the following procedure should be follow- 
ed: 

(1) With the pole and tracing arms approximately at right-angles, 
place the tracing point in the centre of the area to be measured. 

(2) Approximately circumscribe the area, to judge the size of the 
area compared with the capacity of the instrument. If not poss- 
ible the pole should be placed elsewhere, or if the area is too 



AREAS 



479 



large it can be divided into sections, each being measured sep- 
arately. 

(3) Note the position on the figure where the drum does not record 
- this is a good starting point (A). 

(4) Record the reading of the vernier whilst the pointer is at A. 

(5) Circumscribe the area carefully in a clockwise direction and 
again read the vernier on returning to A. 

(6) The difference between the first and second reading will be 
the required area. (This process should be repeated for ac- 
curate results). 

(7) Some instruments have a variable scale on the tracing arm to 
give conversion for scale factors. 

(2) When the fulcrum is inside the figure (Fig. 9.33) 




Fig. 9.33 Planimeter fulcrum inside the figure 

In this case the bar traces out the figure (A T ) abcdefgha - the area 
of the circle (4) A BCDEFGHA; it has rotated through a full circle, 
i.e. a = 2 77. 

A T -A C = l W + l 2 (l- k\2TT 

.: A T = Iw + f (j- k\2n+ A c 

= l w + l 2 (l- }S 2n + -nb 2 (where b = 04) 

= Iw + n{b 2 +fa-2k)] (9.47) 



480 



SURVEYING PROBLEMS AND SOLUTIONS 



This is explained as follows Fig. 9.34: 
If the pointer P were to rotate 

without the wheel W moving, the 

angle OWP would be 90° 
The figure thus described is 

known as the zero circle of radius r; 




Fig. 9.34 Theory of the zero circle 



or 

r 2 



= b 2 -(klf 

= OW z +(l-klf 

= b z - (ktf + I 2 - 2kl 2 + (klf 

= b 2 + I 2 (1 -2k) 



In Eq. (9.47), 



An 



= lw + nr 
= Iw + the area of the zero circle 
The value of the zero circle is quoted by the manufacturer. 



(9.48) 



(9.49) 



N.B. (1) A T - A c = lw. If A c > A T then lw will be negative, i.e. 
the second reading will be less than the first, the wheel 
having a resultant negative recording. 
(2) The area of the zero circle is converted by the manufactur- 
er into revolutions on the measuring wheel and this con- 
stant is normally added to the recorded number of revolu- 



Example 9.12 






/irp ^> ™-o 


1st reading 


3-597 




2nd reading 


12-642 




Difference 


9-045 




Constant 
Total value 
1st reading 


23-515 




32-560 


Arp < Aq 


6-424 




2nd reading 


3-165 




Difference 
Constant 


-3-259 




23-515 




Total value 


20-256 



AREAS 481 

9.3 Plan Areas 



9.31 Units of area 



lsq foot (ft 2 ) = 144 sq inches (in 2 ) 

lsqyard(yd 2 ) = 9 ft 2 

1 acre = 4 roods 

= 10 sq chains = 100 000 sq links 

= 4840 yd 2 = 43 560 ft 2 

1 sq mile = 640 acres 

Conversion factors 

lin 2 = 6-4516 cm 2 1 cm 2 = 0-155000 in 2 

lft 2 = 0-092 903 m 2 

lyd 2 = 0-836 127 m 2 1 m 2 = 1-19599 yd 2 

1 sq chain = 404-686 m 2 

1 rood = 1011-71 m 2 

1 acre = 4046-86 m 2 1 km 2 = 247-105 acres 

0-404686 hectare (ha) 1 ha = 2-47105 acres 

lsq mile = 2-58999 km 2 

= 258-999 ha 

N.B. The hectare is not an S.I. unit. 

The British units of land measurement are the Imperial Acre and the 
Rood (the pole or perch is no longer valid). 

The fractional part of an acre is generally expressed as a decimal 
although the rood is still valid. 

Thus 56-342 acres becomes 

56-342 acres 
_4_ 

56 acres 1-368 roods 

The use of the Gunter chain has been perpetuated largely because 
of the relationship between the acre and the square chain. 
Thus 240362 sq links = 24-036 2 sq chains 

= 2-40362 acres 

The basic unit of area in the proposed International System is the 
square metre (m 2 ). 



482 SURVEYING PROBLEMS AND SOLUTIONS 

9.32 Conversion of planimetric area in square inches into acres 

Let the scale of the plan be — . 

x 

i.e. lin. = xin. 

lsqin. = x 2 sqin. 

v-2 



12 x 12 x 9 x 4840 



acres 



Example 9.13 

Find the conversion factors for the following scales, (a) 1/2500 
(b) 6 in. to 1 mile, (c) 2 chains to 1 inch. 

2500 2 

(a) lin 2 = — — — — - — — — = 0-995 acres 

12 x 12 x 9 x 4840 9 

(4026-6 m 2 = 0-4026 ha) 

(b) 6in.tol.nile Q^ . 

1760x36 1Ar ^ n 

i.e. lin. = = 10 560 in. 

6 

10560 2 

= 17-778 acres 



144x43560 



(71945 m 2 = 7-1945 ha) 

Alternatively, 6 in. = 1 mile 

36 in =1 mile 2 = 640 acres 

. 2 640 

1 in 2 = — — = 17-778 acres 
36 

(c) 2 chains to 1 inch. 

lin. = 200 links 

1 in 2 = 40 000 sq links 

= 0-4 acres (1618-7 m 2 = 0-16187 ha) 
Alternatively, 2 chains to 1 inch (1/1584) 

lin. = 2x66 ft = 132x12 = 1584ft 

1 • 2 1584 2 

1 in* = . = 0-4 acre 

144x43560 

9.33 Calculation of area from co-ordinates 

Method 1. By the use of an enclosing rectangle (Fig. 9.35) 



AREAS 



483 



B(x&)W 




Ofx»*J 



1 



E(x B *) 



Fig. 9.35 Calculation of area by enclosing rectangle 

The area of the figure ABCDE = the area of the rectangle VWXYZ 
-{MVB + BWXC + ACXD + ADYE + MEZ I 

This is the easiest method to understand and remember but is 
laborious in its application. 

Method 2. By formulae using the total co-ordinates 

Applying the co-ordinates to the above system we have: 

Area of rectangle VWXYZ = (x A -xJ(y 2 -y g ) 
of triangle AVB = l(x 2 -x,)(y 2 -y 1 ) 

of trapezium BWXC = - (y 2 - y 3) { (x 4 - x-j) + (x 4 - x 3 ) \ 

of triangle CXD = I(*-y 4 )(*-*) 

DYE = l(y 4 -*)<*,-*) 

AEZ = £<*-*)(*-*,) 

i.e. A = (x 4 )2 - x 4 y 8 - x,^ + x,y 5 ) - i[ x^ - x^, - x,)*, + x,y, + 2x 4 y2 

- 2x 4 y 3 - x z y 2 + x 2 y 3 -x a y 2 + x 3 y 3 
+ x 4 y 3 - x 4 y A -x 3 y 3 + x 3 y 4 + ^ y 4 

- *4 y 5 " x 5 y 4 + x s y 5 + Xgy, - x^ 6 

-*iy,+*,yj 

•*• 4 = i[y,(x 2 -x 6 ) + y 2 (x3-x ) ) + y 3 (x 4 -x 2 ) + y 4 (x B -x 3 ) + y 5 (x 1 -x 4 )] 



484 



SURVEYING PROBLEMS AND SOLUTIONS 



This may be summarised as 



A = £2&<* f 



^-i) 



(9.50) 



i.e. Area = half the sum of the product of the total latitude of each 
station x the difference between the total departures 
of the preceding and following stations. 

This calculation is best carried out by a tabular system. 



Example 9.14 

The co-ordinates of the corners of a polygonal area of ground are 
taken in order, as follows, in feet: 

A (0, 0); B (200, -160); C (630, -205); D (1000, 70); 
E (720, 400); F (310, 540); G (-95, 135), returning to A. 

Calculate the area in acres. 

Calculate also the co-ordinates of the far end of a straight fence 
from A which cuts the area in half. 

To calculate the area of the figure ABCDEFG the co-ordinates 
are tabulated as follows: 

(a) (b) (c) (d) (c)-(d) (a)x{(c)-(d)} 

Station T.Lat. T.Dep. Preceding Following + 



A 
B 
C 
D 
E 
F 
G 





-160 

-205 

70 

400 

540 

135 




200 
630 
1000 
720 
310 
-95 



Dep. 

200 
630 
1000 
720 
310 
-95 




Dep. 

-95 

200 
630 
1000 
720 
310 



295 

630 

800 

90 

-690 

-815 

-310 



6300 



100800 
164000 

276 000 

440 100 

41850 



6300 1022750 
6 300 



2 ) 1016 450 

508 225 ft 2 

.-. Total Area = 508225 ft 2 (47 215-63 m 2 ) 

= 11-667 acres (4-72156 ha) 

From a visual inspection it is apparent that the bisector of the 
area AX will cut the line ED. 

The area of the figure A BCD can be found by using the above 
figures. 



AREAS 



485 



A 
B 
C 
D 



200 1000 -800 

160 200 630 630 

205 630 1000 200 800 

70 1000 630 -630 



100800 

164000 

44100 

2)308900 

154450 ft 2 





J^310,540) 






/ ^^^«f(720,4OO) 




6 <f 






<-95,135)\ 


,''''' 


* {1O00,70) 


A 






(0,0) 


B <S 





(200,-160) 



(630,-205) 
Fig. 9.36 

The area of the triangle AXD must equal 

^(508 225) - 154 450 = 99 662-5 ft 2 
Also area = ^AD.DX sin D 



DX = 



2x99662-5 



AD sin D 
To find length AD and angle D. 

Bearing DA = tan -1 -1000/ -70 = S 85°59'50" W = 265°59'50" 

Length DA = 1000/sin85°59'50" = 1002-45 ft 

Bearing DE = tan" 1 - 280/330 = N 40°18'50" W = 319°41'10" 

Angle 

2x99662-5 



D 
DX 



53°41'20' 



1002-45 sin 53°4l'20 



■„ = 246-76 ft 



To find the co-ordinates of X. (N 40°18'50" W 246*76 ft). 

E x = 840-35 ft 



sin bearing 0*64697 
cos bearing 0*762 51 



E D = 1000*00 
AE = -159*65 
AN = +188*16 



N n = 



70-00 



N, 



258 -16 ft 



486 



SURVEYING PROBLEMS AND SOLUTIONS 



Ch 


eck on Area 








A 








200 


840-35 


-640-35 


B 


-160 


200 


630 





630 100800 


C 


-205 


630 


1000 


200 


800 164000 


D 


70 


1000 


840-35 


630 


210-35 14724-5 


X 


258-16 


840-35 





1000 


- 1000 258 160 
522960 
14724-5 

508 235-5 ft 2 



Method 3. By areas related to one of the co-ordinate axes (Fig. 9.37) 




Fig. 9.37 Areas related to one axis 

Area ABCDE = (bBCc + cCDd + dDEe) - (bBAa + aAEe) 
Using the co-ordinates designated, trapeziums 



( bBAa = '-(x 2+ x,)(y 2 -y,) 
aAEe = Ux, +*,)(y, -y 5 ) 



bBCc = |(x 2+ x 3 )(y 2 -y 3 )' 
cCDd = ^(x 3 +x A )(y 3 -y A ) , -{ 

dDEe = ^(* 4 +*5)(y 4 -y 5 ) t 

i- e. j[ (*2M> - x 2 y 3 + x 3 y 2 - x^) + (x^ - x 3 y A + x A y 3 - x+y A ) 
+ (x A y A - xjs + %y A - a^y 3 ) - (x,y 2 - x,y, + x,^ - x,y, ) 

-O^-^ys+^y,-*.^)] 



(9.51) 



Uyi(x 2 -xJ + y 2 (Xa -x t ) + 3a (x A -x 2 ) + y^-xj + ^(x, -x A )] 



AREAS 



487 



as before 
Area 



^y n (x n+1 -^_,) 



(Eq.9.50) 



Method 4. Area by 'Latitudes* and 'Longitudes* (Fig. 9.38) 

Here Latitude is defined as the partial latitude of a line 

Longitude is defined as the distance from the y axis to the 
centre of the line. 




Fig. 9.38 Calculation of area by latitudes and longitudes 

From Eq. (9.51), 

A = \l\{x 2 + x 3 )iy z -y 3 )^{x 3 + x A ){y 3 -y A ) + {x A +x^(y A -y z )\ 

- {(^ + x,)(y 2 -y l ) + (x, + x 5 )(y 1 -y 5 )}] 

= l[(x, +x z )(y, -y 2 ) + (x 2 + x 3 )(y 2 -y 3 ) + (x 3 + x A )(y 3 -y 4 ) 

+ (* 4 + * 5 ) (y 4 - y 5 ) + (* 5 + *i> (y 8 " ?f > 1 
= -;s(x n + v ,)(yn-yn + ,) ( 952 > 

where ^(x n + x n+1 ) = the longitude of a line. 

Ok - yn+i) = the partial latitude of a line, i.e. the latitude. 

N.B. (1) It is preferable to use double longitudes and thus produce 
double areas, the total sum being finally divided by 2 
(2) This method is adaptable for tabulation using total depar- 
tures and partial latitudes or vice versa. 



488 SURVEYING PROBLEMS AND SOLUTIONS 

Example 9.15 

From the previous Example 9.14 the following table is compiled. 



Stn. 


T. Dep. 


P. Lat. 


Sum of 


Double 


Area 








Adj. T. Dep. + 


- 


A 

















-160 


200 




32000 


B 


200 














-45 


830 




37 350 


C 


630 














+ 275 


1630 


448 250 




D 


1000 














+ 330 


1720 


567600 




E 


720 














+ 140 


1030 


144 200 




F 


310 














-405 


215 




87075 


G 


-95 














-135 


-95 


12825 




A 















1172875 


156 425 










156425 












2)1016 450 












508 225 ft 2 





9.34 Machine calculations with checks 

Using Eq. (9.52), 





(1) 


(2) 


(3) 


(4) 


(5) = (3)x(4) 


(6) 


(7) 


(8) = (6)x(7) 




y 


X 


dy 


Ix 




dx 


Sy 




A 








-160 


200 


- 32 000 


200 


-160 


- 32 000 


B 


-160 


200 


- 45 


830 


- 37 350 


430 


-365 


-156 950 


C 


-205 


630 


+ 275 


1630 


448 250 


370 


-135 


-49 950 


D 


70 


1000 


+ 330 


1720 


567 600 


-280 


470 


-131600 


E 


400 


720 


+ 140 


1030 


144 200 


-410 


940 


- 385 400 


F 


540 


310 


-405 


215 


-87 075 


-405 


675 


-273 375 


G 


135 


-95 
+ 2860 


-135 


-95 


+ 12825 


95 


135 
+ 2220 


+ 12 825 




+ 1145 


+ 745 


+ 5625 


+ 1172875 


+ 1095 


+ 12 825 




-365 
+ 780 


-95 

+ 2765 


-745 


-95 
+ 5530 


-156 425 


-1095 


-660 


- 1 029 275 




+ 1 016 450 




1560 


1016 450 




(X2) 


(X2) 
















1560 


5530 






508 225 fl 2 






508 225ft 2 



AREAS 489 

N.B. (1) From the total co-ordinates in columns 1 and 2, the sum 
and difference between adjacent stations are derived. 

e.g. dy AB = (-160-0) = -160 Zy AB = -160+0 = -160 
dy B c= (-205+160) = - 45 Sy sc = -160-205 = -365 

(2) All the arithmetical checks should be carried out to prove 
the insertion of the correct values. 

(a) 2 x the sum of the total latitudes = Sy (col. 7). 

(b) 2 x the sum of the total departures = Sx (col. 4). 

(c) The Algebraic sum of columns 3 and 6 should equal zero. 

(3) The product of columns 3 and 4 gives column 5, i.e. the 
double area: 

Also the product of columns 6 and 7 gives column 8, i.e. 

the double area: 

Columns 5 and 8 when totalled should check. 

Alternative method 

From the previous work it is seen that the area is equal to one- 
half of the sum of the products obtained by multiplying the ordinate 
(latitude) of each point by the difference between the abscissae 
(departure) of the following and preceding points. 

i- e - A = ^iW ~ *>y, + x 3 y 2 - x, y 2 + x A y 3 - x z y, 



2^3 



+ x 5 y A - x& A + x, yg - x 4 y 5 ) (9. 53) 

This may be written as, 

A = L 



1 v ^r* 2 - 



:i 



(9.54) 



This is interpreted as 'the area equals one half of the sum of the pro- 
ducts of the co-ordinates joined by solid lines minus one half of the 
sum of the products of the co-ordinates joined by dotted lines, 

This method has more multiplications but only one subtraction. 

Using the previous example; 



A = a 



TO.^ 200^ 630^ 1000^ 720 310^ _ 95"] 
2 U^-160' > >-205' > ^ 70 ""400^^540^^135 J 



= ^[-182700 -833750] 



= -[ 1 016 450 ] (negative sign neglected) 



2 

= 508 225 ft 



490 SURVEYING PROBLEMS AND SOLUTIONS 

9.4 Subdivision of Areas* 

9.41 The subdivision of an area into specified parts from a point on 
the boundary (Fig. 9.39) 

B 




Fig. 9.39 Subdivision of an area from a point on the boundary 

Consider the area ABCDE to be equally divided by a line start- 
ing from X on the line ED. 

(1) Plot the co-ordinates to scale. 

(2) By inspection or trial and error decide on the approximate line 
of subdivision XY. 

(3) Select a station nearest to the line XY, i.e. A or B. 

(4) Calculate the total area ABCDE. 

(5) Calculate the area AXE. 

(6) Calculate the area AXY = \ABCDE - AXE. 

(7) Calculate the length and bearing of AX. 

(8) Calculate the bearing of line ED. 

(9) Calculate the length A Y in triangle AYX. 

N.B. Area of triangle AYX = -AX.XY sin X. 

As the area, AX and X are known, AY is calculated: 

AY Area of triangle AYX ,g r« 

-AX sin X 

2 

(10) Calculate the co-ordinates of Y. 

* For a complete analysis of this the reader is advised to consult The Basis 
of Mine Surveying by M.H. Haddock. 



AREAS 



491 



9.42 The subdivision of an area by a line of known bearing (Fig. 9.40) 




Fig. 9.40 Subdivision of an area by a line of known bearing 

Construction 

Given the area ABCDE and the bearing of the line of sub-division 
XY, then EF on the given bearing and XY will be parallel to this, a 
perpendicular distance d away. 

Draw FG perpendicular to XY. 
HX perpendicular to EF. 

The area AYXE = ^area ABCDE 

= AAFE + AFYG+FGXH + AHXE. 

(1) From the co-ordinates the length and bearing of AE can be 
calculated. 

(2) In the triangle AFE the area can be obtained by first solving 
for the length EF. 

(3) The area of the figure FY GX EH can thus be obtained in terms 
of d, i.e. 

triangle FYG = -d cot a 

rectangle FGXH = d(EF - HE) 

= d(EF- dcotB) 

triangle HXE = ^d 2 cot)8 



492 SURVEYING PROBLEMS AND SOLUTIONS 

.'. dEF + irf 2 (cota-cot)3) = Area of AYXE - Area of AAFE. 

(9.56) 
This is a quadratic in d as the angles a and B are found from 
the bearings. 

From the value of d the co-ordinates of F and E can be obtained. 

9.43 The subdivision of an area by a line through a known point 
inside the figure (Fig. 9.41) 




Fig. 9.41 Subdivision of an area by a line through a known 
point inside the figure 

Construction 

Given the area ABC D E and the co-ordinates of the fixed point 
H, join EH and produce to cut AB in G. Assume the dividing line 
XY is rotated <x° about H. 

From the co-ordinates: 

(1) Calculate the length and bearing EH. 

(2) In the triangle AGE calculate the length EG (this gives the 
length AG) and the area. 

(3) The required area AYXE = A AGE - A YGff + AE//X 

= A 4GE- area (A) 

To find the missing area, 

A = \EH.HX sin a +±GH.HY sin a 

! _ EH sin <t> sin a i GH sin 6 sin a 

= |E// x , x + ^Gtf x 

2 sin(a+0) z sin(a+0) 



AREAS 



493 



EH Z 



GH' 



cot a + cot <f> cot a + cot 6 



.] 



24 = 



EH' 



GH' 



• + 



cot a + cot 



(9.57) 



cot a + cot <fi 

As the lengths EH and GH are known, and <f> and are obtain- 
able from the bearings, this is a quadratic equation in cot a, from which 
a may be found, and thus the co-ordinates of X and Y. 

These problems are best treated from first principles based on the 
foregoing basic ideas. 

Example 9.16. In a quadrilateral A BCD, the co-ordinates of the 
points, in metres, are as follows: 

Point E N 

A 

B -893-8 

C +634-8 -728-8 

D +1068-4 +699-3 

Find the area of the figure by calculation. 

If E is the mid-point of AB, find, graphically or by calculation, 
the co-ordinates of a point F, on the line CD, such that the area 
AEFD equals the area EBCF. 
N.B. 



Co-ordinates of E = -(A + B) 



i.e. 



0, ^x -893-8 = 0, 




446-9 

'(1068-4, 699-3) 
26° 06' 



[892-5,119-8] 



(634-8,-728-8) 



Fig. 9.42 



494 SURVEYING PROBLEMS AND SOLUTIONS 

x y dx iy dy ix 

A -893-8 - -893«8 104 742-00 

B -893-8 634-8-1622-6-1030 026-48 165-0 634-8 2432339-92 

C 634-8 -728-8 433-6 -29-5 -12 791-20 1428-1 1703-2 

D 1068-4 +699-3 - 1068-4 699-3 -747 132-12 -699-3 1068-4 -747 132-12 

1703-2 699-3 1068-4 -699-3 1593-1 3406-4 2537081-92 

-1622-6 -1068-4 -2545-9 -1789 949-80 -1593-1 -747 132-12 

1703-2 -923-3 - 1846-6 - 1 789 949-80 3406-4 +1789 949-80 

3406-4-1846-6 

Area = 894 974-9 it 2 

Referring to Fig. 9.42, 

. 1068-4 - 

Bearing ED = tan" 1 —— -— - = tan -1 0-93212 = N 42°59'20"E 

& 699-3 + 446-9 

Length ED = 1068-4 sin 42°59'20" = 1566-9 

In triangle ADE, Area = -AE.ED sin 42°59'20" 

= i x 446-9 x 1566-9 sin42°59'20" 

= 238735-4 m 2 
-•- Area triangle EDF = 894 974'9 _ 238 735-4 sq ft = 208752 m 2 

, 634-8 - 1068-4 
Bearing DP = Bearing DC = tan ,728-8 -699-3 

= tan"' 0-30362 = S le^^O" W 
Angle ADF = 42°59'20" - 16°53'20" = 26 o 06 , 

Using Eq. (9.55), 

Area A EDF _ 2x208752 

DF = ^-ED sin EDF = 1566 ' 9 sin 26 ° 06 ' = — " 
2 

To obtain the co-ordinates of F, 

Line DF S 16°53'20" W Length 605-66 m 
P. Dep. 605-66 sin 16°53'20" = -175-9 
P. Lat. 605-66 cos 16°53'20" = -579-5 
T. Dep. of F 1068-4 - 175-9 = 892-5 m 

T. Lat . of F 699-3-579-5 = 119-8 m 

Example 9.17 

With the previous co-ordinate values let the bearing of the divid- 
ing line be N 57°35'10" E. 



AREAS 



495 



Construction 

Draw line AG on this bearing 
and EF parallel to this a perpend- 
icular distance d away. 

. 1068-4 
Bearing AD = tan 599.3 

= tan" 1 1-52781 
= N 56°47'40" E 
Length AD = 1068-4 cosec56°47 '40 
= 1276-9 m 




Fig. 9.43 



In triangle ADG, 

A = 57°35'10" -56°47'40" = 0°47'30" 

D = 56°47'40" - 16°53'20" = 39°54'20" 

G = 180 -(57 o 35'10" - 16°53'20") = 139°18'10" 

180°00 , 00" 
AD sin D 1276-9 sin 39°54'20" 



AG 



sin 6* 



sin 139°18'10" 



= 1256-3 m 



Area triangle ADG = ^AD.AG sin A 

= jx 1276-9 x 1256-3 sin0°47'30" 
= 11084-8 m 2 

Area AGFE = ± Area ABCD -A ADG 

2 

= 447487-5-11084-8 = 436402-7 m a 



In figure AGFE, Area = AA7E + AHFJ + AFHG 



1 J2 



' j2 



= ± d z cot £ + d( AG -d cot a) + -d* cot a 



= 1256-3 d + \ d\cot j8 - cot a) 



where a = 57°35'10" - 16°53'20" = 40°41'50" 

j8 = 57°35'10" 
/. 436 402-7 = 1256-3 d- 0*263 88 d 2 



496 SURVEYING PROBLEMS AND SOLUTIONS 

This is a quadratic equation in d 

0-263 88 d z - 1256-3 d+ 436402-7 = 



d = 



1256-3 ± N /(1256-3 2 - 4 x 0-26388 x 436 402-7) 
2x0-26388 

1256-3 ±^(1578 289-7 - 460631-8) 
2x0-26388 

1256-3 ±V 1117657-9 
2x0-26388 

1256-3 + 1057-2 2313-5 199-1 

or 



2x0-26388 2x0-26388 2x0-26388 

The first answer is not in accordance with the data given. 

.-. d = 377-26 m 
/. Co-ordinates of E = 0, and -377-26 cosec 57°35'10" 
= Total Pep. 

Total Lat. -446-9 m 



Length EF = AG - HG + EJ 

= 1256*3- 377*26 cot a +377*26 cot j8 

= 1256-3 + 377-26 (cot )8- cot a) 

= 1256-3 - 199-1 = 1057-2 

Co-ordinates of F: 

P. Dep. 1057-2 sin 57°35'10" = +892-5 m 

P. Lat. 1057-2 cos 57°35'10" = +566*7 m 

Total Dep. of F = + 892*5 = +892-5 m 

Total Lat. of F = -446*9 + 566-7 = +119-8 m 

Example 9.18 

Given the previous co-ordinate values let the dividing line pass 
through a point whose co-ordinates are (703*8, 0). 

From previous information, 

AD is N 56°47'40" E, 1276*9 m 
DC is S 16°53'20" W 

AD sin D 1276*9 sin 39°54'20" ocr _ 

In triangle ADG, AG = ^- = ; — - — = 855-9 m 

sinG sinl06°53'20" 



AREAS 



497 




39° 54' 20" 



<J>=106 , 53'20" 



Fig. 9.44 
AH = 703-8 (due E) 
HG = 855-9 - 703-8 = 152-1 m 



Area = -AD. AG sin A 



= \ x 1276-9 x 855-9 sin 33°12'20" = 299 268 m 2 



Now Area ADFE = ~ Area ABCD 

= AADF+ AAHE- AHFG 
: &AHE-AHFG = 447488-299 268 



i.e. by Eq.9.57, 148 220 = 



AH' 



- 447 488 m 2 

= 148 220 m' 
HG 2 



2(cot a + cot 6 ) 2(cot a + cot <f> ) 

As 6 = 90°, 

296440 cot a(cot a + cot <f>) = AH 2 (cota+ cot0) - HG 2 cota 
i.e. 296440cot 2 a+cota[296 440cot<£-AH 2 +HG 2 ]-4tf 2 cot<£ = 
thus 296440 cot 2 a- 562 200 cot a + 150 388 = 

Solving the quadratic gives a = 32°25' 

The co-ordinates of E are thus 0, and 703-8 tan32°25' 
i.e. (0, -446-9 m) 



Exercises 9 

1. In the course of a chain survey, three survey lines forming the 
sides of a triangle were measured as follows: 



498 SURVEYING PROBLEMS AND SOLUTIONS 

Line Length (links) Inclination 

AB 570 level 

BC 310 1 in 10 

CA 495 7° 

On checking the chain after the survey, it was found that its length 
was 101 links. 

Calculate the correct plan area of the triangle. 

(E.M.E.U. Ans. 0-77249 acres) 

2. A piece of ground has been surveyed with a Gunter's chain with 
the following results in chains: AB 11*50, CA 8-26, DB 10-30, 

CE 12-47, BC 12-20, CD 9-38, DE 6-63. Calculate the area in acres. 
Subsequently it was found that the chain was 0-01 chain too long. Find 
the discrepancy in the previous calculation and indicate its sign. 

(L.U. Ans. 12-320 acres; -0*248 acres) 

3. Undernoted are data relating to three sides of an enclosure, AB, 
BC and CD respectively, and a line joining the points D and A. 



Line 


Azimuth 


Length (ft) 


AB 


010°00' 


541-6 


BC 


088°55' 


346-9 


CD 


159° 19' 


601*8 


DA 


272°01' 


654*0 



The fourth side of the enclosure is an arc of a circle to the south 
of DA, and the perpendicular distance from the point of bisection of 
the chord DA to the curve is 132*6 ft. Calculate the area of the en- 
closure in acres. 

(Ans. 7*66 acres) 

4. Plot to a scale of 40 inches to 1 mile, a square representing 2H 
acres. By construction, draw an equilateral triangle of the same area 
and check the plotting by calculation. 

(Ans. Side of square 2*5 in. 
Side of triangle 3-8 in.) 

5. The following offsets 15 ft apart were measured from a chain line 
to an irregular boundary: 

23-8, 18-6, 14-2, 16-0, 21*4, 30*4, 29*6, 24*2 ft. 
Calculate the area in acres. 

(Ans. 0*0531 acres) 

6. Find the area in square yards enclosed by the straight line bound- 
aries joining the points ABCDEFA whose co-ordinates are: 



AREAS 499 

Eastings (ft) Northings (ft) 

A 250 75 

B 550 175 

C 700 425 

D 675 675 

E 450 675 

F 150 425 

(R.I.C.S. Ans. 24791-6 yd 2 ) 

7. The following table gives the co-ordinates in feet of points on the 
perimeter of an enclosed area A BCDEF. Calculate the area of the 
land enclosed therein. Give your answer in statute acres and roods. 

Point Departure Latitude 

+ - + 

A 74-7 105-2 

B 63-7 261-4 

C 305-0 74-5 

D 132-4 140-4 

E 54-5 192-4 

F 571-9 108-3 

(R.I.C.S. Ans. 4 acres 0-5 roods) 

8. Using the data given in the traverse table below, compute the area 
in acres contained in the figure A BCD EA. 



Side 


Latitude (ft) 


Departure (ft) 


AB 


+ 1327 


-758 


BC 


+ 766 


+ 805 


CD 


-952 


+987 


DE 


-1949 


+ 537 


EA 


+808 


-1572 



(I.C.E. Ans. 73-3 acres) 

9. State in square inches and decimals thereof what an area of 10 
acres would be represented by, on each of three plans drawn to scale 
of (a) 1 inch = 2 chains (b) 1/2500 and (c) 6 in. = 1 mile. 

(Ans. (a) 25 in 2 ; (b) 10-04 in 2 ; (c) 0-562 in 2 ). 

10. State what is meant by the term 'zero circle' when used in con- 
nection with the planimeter. 

A planimeter reading tens of square inches is handed to you to 
enable you to measure certain areas on plans drawn to scales of (a) 
1/360 (b) 2 chains to 1 inch (c) 1/2500 (d) 6 in. to 1 mile and (e) 
40 in. to 1 mile. State the multiplying factor you would use in each 



500 SURVEYING PROBLEMS AND SOLUTIONS 

instance to convert the instrumental readings into acres. 

(Ans. (a) 0-2066 (b) 4-0 (c) 9-96 
(d) 177-78 (e) 4-0) 

11. The following data relate to a closed traverse: 



Line 


Azimuth 


Length (m) 


AB 


241°30'00" 


301-5 


BC 


149°27'00" 


145-2 


CD 


034°20'30" 


415-7 


DE 


079°18'00" 


800-9 


Calculate (a) the 1 


ength and beari 


tig of the line £ 



est 30", 

(b) the area of the figure A BCDEA, 

(c) the length and bearing of the line BX which will 
divide the area into two equal parts, 

(d) the length of a line XY of bearing 068°50' which 
will divide the area into two equal parts. 

(Ans. (a) 307°54' ; (b) 89 290 m 2 ; (c) 526-4 m 
091°13'48"; (d) 407-7 m) 

Bibliography 

HADDOCK, M.H., The Basis of Mine Surveying (Chapman & Hall). 

CLENDINNING, J., The Principles of Surveying (Blackie). 

McADAM, R., Colliery Surveying (Oliver and Boyd). 

MIDDLETON, R.E. and CHADWICK, o., A Treatise on Surveying (Spon). 

THOMAS, W.N., Surveying (Edward Arnold). 

SALMON, V.G., Practical Surveying and Fieldwork (Griffin). 

PARRY, R. and JENKINS, W.R., Land Surveying (Estates Gazette). 

DA VIES, R.E., FOOTE, F.s. and KELLY, J.W., Surveying: Theory and 

Practice (McGraw-Hill). 

MINISTRY OF TECHNOLOGY, Changing to the Metric System (H.M.S.O.). 

SMIRNOFF, M.V., Measurements for Engineering and Other Surveys 

(Prentice-Hall). 



10 VOLUMES 



10.1 Volumes of Regular Solids 

The following is a summary of the most important formulae. 
Prism (Fig. 10.1) 

V = cross-sectional area x perpendicular height 
i.e. V = Ah = A y h, (10.1) 



Fig. 10.1 



a b 

* 1 — V 

Hi 




(a) 



(b) 



Cylinder (Fig. 10.2) 

This is a special case of the prism. 

V = Ah = 7Tr 2 h 



(10.2) 





A = 1/4 rrab 



(a) 

Fig. 10.2 

In Fig. 10.2 (b) the cylinder is cut obliquely and thus the end area 
becomes an ellipse, i.e. V = Ah. 



V = —rrabh 
4 



(10.3) 



= -^ Troth (as b = 2r) 
= v4,/i, = nr z h % 



(10.4) 
(10.5) 



502 SURVEYING PROBLEMS AND SOLUTIONS 

Pyramid (Fig. 10.3) 

V = — base area x perpendicular height 



= 3 Ah 



(10.6) 




A — area of rectangle A « area of triangle 

Fig. 10.3 

Cone (Fig. 10.4) 

This is a special case of the pyramid. 

V = \-Ah = \iTT z h 
3 3 



(10.7) 




A = JT/-2 (circle) A = 'A rrao( ellipse) 



Fig. 10.4 
In Fig. 10.4(b) the base is in the form of an ellipse. 



V = — nabh 



= -T-rrarh (as fe = 2r) 



(10.8) 
(10.9) 



Frustum of Pyramid (Fig. 10.5) 

V = ^(A + B + y/M) 

where h = perpendicular height 

A and B = areas of larger and smaller ends respectively. 



(10.10) 



VOLUMES 



503 




Fig. 10.5 



Frustum of Cone (Fig. 10.6) 

This is a special case of the 
frustum of the pyramid in which 
A = ttR 2 , B = TTr*. 



•'• V = ^bTR Z +Trr 2 +y/7TR Z 7Tr Z ] 

= y[K 2 +r 2 +/?r] (10.11) 



Wedge (Fig. 10.7) 



Fig. 10.7 





f^ J 



504 



SURVEYING PROBLEMS AND SOLUTIONS 



V = Sum of parallel edges x width of base x -?■ perpendicular 
height. ° 



i.e. 



wh 



(x + y + z) 



(10.12) 



The above formulae relating to the pyramid are proved as follows 
(Fig. 10.8). 



Area 6 



Area of 
midsection M 




Area A 



Fig. 10.8 

Let A = the base area 

h A = the perpendicular height 

B = the area of any section parallel to the base and at a perpen- 
dicular distance h a from the vertex. 



Then 




8V = 


B8h 








but 




A 
B = 


h A 

h% 




.-. 


B 


Ah% 

n A 






.'. V = 


A 


h A 

J 


h% 


dh B 








= 


A 

#A 


X 


hi 
3 


1 

" 3 


Ah A 


In the 


case 


of the frustum (h 


A~ 


h B 


= 


h), 





(Eq. 10.6) 



VOLUMES 



505 



V -£J* 



dhi 



A 



= TJT iK ~ h B^(- h A + h A h B + h B ) 



3/z 

h 
3 



A + 



Ah, 



Ah\ 



(10.13) 



But 



hi 



B = 



Ahl 



and 



y/B 

y/A 



•'• V = -[A + y/AB + B] (Eq. 10.10) 

If C is the area of the mid-section of the pyramid, then 



£__V27 = ^_ 

A h A 4h% 

.'. A = 4C 

. .. Ah A (A + 4C)h A 

•■ V = -~ = — -^ (10.14) 

o O 

Similarly, if M is the area of the mid-section between A and B, then 
M _ ii(h A + h B )\ 2 _ (h A + h B f 



. 4M = A(h*+2h A h B + h%) 



h A Ha 



= A + 24^ + B 
-r- = 1(4M- 4 - B) 



506 



SURVEYING PROBLEMS AND SOLUTIONS 



Substituting this value in Eq. (10.13), 



V = -[A + 4M+B] (10.15) 

The Prismoidal Formula 

From Eq. 10.15 it is seen that the volumes of regular solids can be 
expressed by the same formula, viz. the volume is equal to the sum of 
the two parallel end areas + four times the area of the mid-section x 
1/6 the perpendicular height, i.e. 



= ^ [A + AM + B] 
6 



(Eq. 10.15) 



This formula is normally associated with the prismoid which is 
defined as 'a solid having two parallel end areas A and B, which may 
be of any shape, provided that the surfaces joining their perimeters are 
capable of being generated by straight lines.' 

N.B. The mean area is derived from the average of the corresponding 
dimensions of the two end areas but not by taking the average of A 
and B. *?** * 

■V 



Fig. 10.9 




Area M 



Area B 



VOLUMES 



507 



Newton's proof of this formula is to take any point X on the mid- 
section and join it to all twelve vertices of the three sections, Fig. 10.9. 
The total volume then becomes the sum of the ten pyramids so formed. 

This formula is similarly applicable to the cone and sphere. 



The cone (Fig. 10.10) 

= 2irr 2 h 



= -7rr 2 h (Eq.10.7) 



Area 6 



Area 



Area A*trr' 




"-■«/? 



Fig. 10.10 



The sphere (Fig. 10.11) 

V = ^[0 + 477r 2 +0] 
6 



Area B m 



= r r 



(10.16) Area M + 
- rrr z 



Area A—0 




Fig. 10.11 

N.B. The relative volumes of a cone, sphere and cylinder, all of the 
same diameter and height, are respectively 1, 2 and 3, Fig. 10.12. 

2r 
Cone = — x rrr 2 
3 



= -nr' 
3 



(10.17) 



Sphere = -ttt* (Eq. 10.16) 

Cylinder = 2r x nr z 

= 2rrrf (10.18) 




Fig. 10.12 



508 SURVEYING PROBLEMS AND SOLUTIONS 

Applying the prismoidal formula to the frustum of a cone, 

-![-*w + *,(*-f] 



V 



77" h 

= — [R 2 +r z +R 2 +2Rr+r 2 ] 
6 



rrh 



[R 2 +r 2 +Rr] 



(Eq. 10.11) 



Applying the prismoidal formula to the wedge, 

v -6l2 ( * +y) + 4 (-r + _ r)4 +0 J 

wh[ x y x z z y *J 
= TL2 + 2 + 2 + 2 + 2 + 2j 



— ix + y + z\ 
6 



(Eq. 10.12) 



It thus becomes very apparent that of all the mensuration formulae 
the PRISMOIDAL is the most important. 

If in any solid having an x axis the areas (A) normal to this axis 
can be expressed in the form 

A = bx 2 + ex + d 

then the prismoidal formula applies precisely. 

The sphere may be regarded as made up of an infinite number of 

small pyramids whose apexes meet at the centre of the sphere. The 

heights of these pyramids are then equal to the radius of the sphere. 

r 
Volume of each pyramid = area of base x - 

x 
Volume of sphere = surface area of sphere x — 



Surface area of sphere = 

Sector of a sphere (Fig. 10.13) 

This is a cone OAC with a sphe- 
rical cap. ABC. 

The volume can be derived from 
the above arguments: 

Volume of sector = (curved surface 
area of segment) x r/3 



volume of sphere 
l/3r 



= 4nr 2 (10.19) 




VOLUMES 509 



(10.20) 



(by Eq.9.33) = Irrrxhx^ 

2 

= —nr z h 
J 

Segment of sphere 

This is the sector less the cone OAC 

V = —7rr 2 h - r^Tw 2 (r - h) 

But w z = r 2 -(r-/i) 2 = r z -r z +2rh-h z 

2 , 1 

V = -7rr z h- -7T(2rh-h z )(r-h) 

2 2 1 

= ^-7Tr z h -—7Tr 2 h + irrh 2 — nh 3 

3 3 3 

= -7Th z (3r-h) (10.21) 

10.2 Mineral Quantities 

Flat seams 

The general formula for calculating tonnage is: 

plan area (ft 2 ) x thickness (ft) x 62'5 x S.G. 
Tonnage - — tons 

ZZW (10.22) 

Here 62-5 ~ the weight of 1 ft 3 of water in pounds 

S.G. = the specific gravity of the mineral. 

This gives the tonnage in a seam before working and takes no 
account of losses. 

Taking coal as a typical mineral, alternative calculations may be 
made. 

Tonnage = plan area (acres) x thickness (in.) x 101 S.G. 

(10.23) 
Here 1 acre of water 1 inch thick weighs approximately 101 tons. 
When the specific gravity of coal is not known, either 

(a) Assume S.G. of 1-25 - 1*3. 

(b) Assume 125 tons per inch/ acre 

1250 - 1500 tons per foot/ acre. 

(c) Assume 1 yd 3 of coal weighs 0*9 - 1*0 ton, or 

(d) Assume 1 ft 3 of coal weighs 80 lb. 

For loss of tonnage compared with the 'solid' estimate assume 
15-20%. 



510 



SURVEYING PROBLEMS AND SOLUTIONS 



Based on the International System (S.I.) units, the following con- 
version factors are required: 



lton = 1016-05 kg 
1 cwt = 50-802 3 kg 
1 lb = 0-453 592 37 kg 



1 ft = 0-3048 m 

1 ft 2 = 0-092 903 m 2 

1 acre = 4046-86 m 2 

1 ft 3 = 0-028 316 m 3 

lyd 3 = 0-764 555 m 3 

The weight of water is 1 g/cm 3 at 4°C (i.e. 1000 kg/m 3 ) 

/. Coal weighs ^ 1250 - 1300 kg/m 3 ( x 1000 kg/yd 3 ) 

1 gallon = 4-546 09 litres = 0-004 546m 3 

Inclined seams 

The tonnage may be obtained by using either (a) the inclined area 
or (b) the vertical thickness, i.e. 

V = A.fseca (10.24) 

where V = plan area 

t = thickness 

a = angle of inclination of full dip of seam 



Example 10.1 In a pillar and stall, or stoop and room workings, the 
stalls or rooms are 12 ft in width and the pillars are 40 yd square. 
Calculate the approximate tonnage of coal extracted from the stalls or 
rooms, in a seam 7 ft 9 in. in thickness, dipping 19° from the horizon- 
tal, under a surface area V/2 acres in extent. Assume a yield of 125 
tons per inch-acre, and deduct 3%% for loss in working. 

(M.Q.B./M ) 



1 1 L 



Stall 



120' 






Pillar 



120' 



1 






Fig. 10.14 



VOLUMES 511 



In Fig. 10.14, Total Area = (120 + 12) 2 

Pillar Area = 120 2 

/120\ 2 
.'. % extraction = 10 °-[7^j x 10 ° 



■■»'■- -inr 



(-(i 



= 17-36% 
Plan area of extraction = 1*5 x 17*36/100 acres 
Inclined area of extraction = 1*5 x (17*36/100) x sec 19° 
Volume of coal extracted = 1*5 x (17*36/100) sec 19 x 7*75 x 12 x 125 

= 3201*5 tons 

, , 3201*5 x 15 _ fti 
Loss of volume = --rrr = 120 tons 

Approximate tonnage extracted = 3200 - 120 

= 3080 tons 

Exercises 10(a) (Regular solids) 

1. A circular shaft is being lined with concrete, of average thickness 
18 in. The finished inside diameter is 22 ft and a length of 60 ft is 
being walled. In addition 23 yd 3 of concrete will be required for a 
walling curb. 

If each yd 3 of finished concrete requires (a) 700 lb cement (b) 
1600 lb sand and (c) 2 500 lb aggregate, find, to the nearest ton, the 
quantity of each material required to carry out the operation. 

(Ans. 84 tons cement; 192 tons sand; 300 tons aggregate) 

2. A colliery reservoir, circular in shape, with sides sloping at a 
uniform gradient and lined with concrete is to be constructed on level 
ground to the undernoted inside dimensions: 

Top diameter 40 m 

Bottom diameter 36 m 

Depth 9 m 

The excavation is to be circular, 42 m in diameter, with vertical 
sides 10*5 m deep. 

Calculate the volume of concrete required. 

(Ans. 122*63 m 3 ) 

3. Two shafts — one circular of 20 ft diameter, and the other rectan- 
gular 20 ft by 10 ft -are to be sunk to a depth of 710 yd. The mate- 
rial excavated is to be deposited in the form of a truncated cone with- 
in an area of level ground 100 yd square. If the top of the heap is to 



512 SURVEYING PROBLEMS AND SOLUTIONS 

be level and the angle of repose of the material 35°, what will be the 
ultimate height of the heap with the diameter at its maximum ? Assume 
the proportion of broken to unbroken strata to be 5 to 3 by volume 
(take 77= (22/7)). 

(Ans. 38-9 ft) 

4. An auxiliary water tank in the form of a cylinder with hemispheri- 
cal ends is placed with its long axis horizontal. The internal dimen- 
sions of the tank are (i) length of cylindrical portion 24 m (ii) dia- 
meter 5 m (iii) overall length 29 m. 

Calculate (a) the volume of the tank and 

(b) the amount of water it contains to the nearest 100 
litres when filled to a depth of 1 *07 m . 

(Ans. (a) 536-69 m 3 ; (b) 11400 litres) 

5. Two horizontal drifts of circular cross-section and 16 ft excava- 
ted diameter cross each other at right-angles and on the same level. 
Calculate the volume of excavation in ft 3 which is common to both 

drifts ' (M.Q.B./S Ans. 2731 ft 3 ) 

6. The plan of a certain building on level ground is a square with 
sides 200 ft in length for which mineral support is about to be acqui- 
red. The south side of the building is parallel to the line of strike of 
the seam, the full dip of which is due South at the rate of 12 in. to the 
yard. The floor of the seam is 360 yd under the surface at the centre 
of the building. 

Draw a plan of the building and protecting block to a scale of 1 in 
= 200 ft, allowing a lateral margin equal to one third of the depth of 
the seam at the edge of the protecting block opposite the nearest point 
of the protected area. Thereafter calculate the tonnage of coal contain- 
ed in the protecting block, the seam thickness being 70 in. and the 
sp. gr. 1*26. 

(M.Q.B./S Ans. 182860 tons) 

7. A solid pier is to have a level top surface 20 ft wide. The sides 
are to have a batter of 2 vertical to 1 horizontal and the seaward end 
is to be vertical and perpendicular to the pier axis. It is to be built on 
a rock stratum with a uniform slope of 1 in 24, the direction of this 
maximum slope making an angle whose tangent is 0*75 with the direc- 
tion of the pier. If the maximum height of the pier is to be 20 ft above 
the rock, diminishing to zero at the landward end, calculate the volume 
of material required. 

(L.U. Ans. 160000 ft 3 ) 

8. A piece of ground has a uniform slope North and South of 1 verti- 
cal to 20 horizontal. A flat area 200 ft by 80 ft is to be made by cut- 
ting and filling, the two volumes being equal. Compare the volumes of 



VOLUMES 



513 



excavation if the 200 ft runs (a) North and South (b) East and West. 
The side slopes are to be 1 vertical to 2 horizontal. 

(L.U. Ans. (a) 24 300 ft 3 ; (b) 4453 ft 3 ) 

10.3 Earthwork Calculations 

There are three general methods of calculating volumes, which use 
(1) cross-sectional areas, (2) contours, (3) spot heights. 



10.31 Calculation of volumes from cross-sectional areas 

In this method cross-sections are taken at right-angles to some 
convenient base line which generally runs longitudinally through the 
earthworks. The method is specifically applicable to transport cons- 
tructions such as roads, railways and canals but may be applied to any 
irregular volume. 

The cross-sectional areas may be irregular and thus demand the 
use of one of the previously discussed methods, but in many transport 
constructions the areas conform to various typical shapes, viz. sections 
(a) without crossfall, (b) with crossfall, (c) with part cut and part 
fill, (d) with variable crossfall. 

(a) Sections without crossfall, i.e. surface level (Fig. 10.15) 



n Formation peg 




3 (b) Embankment 



Fig. 10.15 Sections without crossfall 

The sections may be cuttings or embankments but in either case 
the following terms are used: 

Formation width (w) 

Formation height Qi ), measured on centre line (<L) 



514 



SURVEYING PROBLEMS AND SOLUTIONS 



Side width (W) , for the fixing of formation pegs, measured from 

centre line. 
Side slopes or batter 1 in m, i.e. 1 vertical to m horizontal 



w 
Thus W = — + mh 

Cross-sectional area = ^ (w + 2W) 

= -yCw + vv + 2mh ) 
A = h (w + mh Q ) 



(10.25) 



(10.26) 



Example 10.2 A cutting formed in level ground is to have a forma- 
tion width of 40 ft (12- 19 m) with the sides battering at 1 in 3. If the 
formation height is 10 ft (3*05 m) find (a) the side width, (b) the 
cross- sectional area. 



Fig. 10.16 



3x10' 




10' 



20' 



S* 



Here W = 40 f t (12*19 m) 
h = 10 ft (3-05 m) 

m = 3 



w 
W = - + mh 



= 20 + 3 x 10 = 50_ft (15*24 m) 

Area A = h (w + mh ) 

= 10(40 + 30) = 700 ft 2 (65-03 m 2 ) 

The metric values are shown in brackets. 

(b) Sections with crossfall of 1 in k (often referred to as a two- 
level section) 

In both the cutting and embankment the total area is made up of 
three parts, Fig. 10.17. 

(1) Triangle AHB Area = &M, 

(2) Trapezium BHFD Area = h w 

(3) Triangle DFE Area = l /2h 2 d 2 . 



VOLUMES 



515 




* -\ 



(a) Cutting 




(b) Embankment 
Fig. 10.17 Sections with cross fall 

In both figures, h. ~ h Q _ ^ 

2k 



h, = h 



o + 



w 
2fc 



By the rate of approach method(see p. 432) 

ft, h^mk 



<*i = 



1 1 k + m 
m k 



(10.27) 
(10.28) 

(10.29) 



516 SURVEYING PROBLEMS AND SOLUTIONS 

««j j ^2 h 2 mk 

and d 2 = —±- = -i— (10.30) 



(10.31) 



"2 




1 1 

m k 


k - m 


••■ H\ 


= 


w 

2 + d ' 


2 fc + m 


W 2 


= 




2 k - m 


Total area 


= 


~Mi + 


K w + 2 hzdz 



(10.32) 



(10.33) 

The area of the cross-section is best solved by working from first 
principles, but if the work is extensive a complete formula may be re- 
quired. 

Given the initial information as w, h , m and k, substitution of 
these values into the various steps gives, from Eq. (10.33), 

( h °~lk) mk ( h ° + 2r) mk 

A =j t*L — + ± f~ + wh 

2(Jk + m) 2(fc - ro) 

- u — + wh n 



2(k 2 - m 2 ) 
mk[2H 2 k + 2(-) + -£-J 



2(*- - m 2 ) + Who 

. " [*»'*' + Q + ^ m \ ^ (10 .34) 

fc 2 - m z + W ° 

Example 10.3 The ground slopes at 1 in 20 at right-angles to the 
centre line of a proposed embankment which is to be 40 ft (12* 19 m) 
wide at a formation level of 10 ft (3*05 m) above the ground. If the 
batter of the sides is 1 in 2, calculate (a) the side width, (b) the 
area of the cross-section. 
In Fig. 10.18, 

w = 40 ft (12-19 m) 

h = 10 ft (3-05 m) 



VOLUMES 



517 



Then 



771 = 2 

k = 20 




«*i 



d, = 



9 x 2 x 20 360 

= — = 16-36 ft 

20 + 2 22 



11 x 2 x 20 



440 



= 24-44 ft 



20-2 18 

W, = 20 + 16-36 = 36-36 ft (11-08 m) 
W 2 = 20 + 24-44 = 44-44 ft (13*55 m) 

Area = Vilh^ + fe 2 d^] + wh 

= l / 2 [9 x 16-36 + 11 x 24-44] + 40 x 10 
= V 2 [147-28 + 268-88] + 400 
= 608-08 ft 2 (56-49 m 2 ) 

By Eq. (10.34), 

2[100 x 400 + 400 + 40 x 10 x 2] 

^4 _ . . — + 40 x 10 



400 - 4 
40000 + 400 + 800 



198 
= 608-08 ft 2 



+ 400 



or converted into S.I. units, 

2[3-05 2 x 400 + 6-095 2 + 12-19 x 3-05 x 2] 



A = 



396 
[3721 + 37-15 + 74-36] + 37>18 



+ 12-19 x 3-05 



198 
= 19-36 + 37-18 



56-54 m : 



518 SURVEYING PROBLEMS AND SOLUTIONS 

(c) Sections with part cut and part fill (Fig. 10.19) 




Fig. 10. 19 Section part cut/part fill 

As before, the formation width BD = w 
the formation height FG = h 
the ground slope = 1 in k 

but here the batter on the cut and the fill may differ, so 
batter of fill is 1 in n 
batter of cut is 1 in m. 

The total area is made up of only 2 parts: 

(1) Triangle ABC Area = l / 2 h,d, 

(2) Triangle CED Area = y i h z d z 

w w 

<*, = 2 ~ X = 2 " **■ 

w w 

d 2 =- + x = —+kh 



(10.35) 



(10.36) 



By the rate of approach method and noting that /i, and h z are now 
required, it will be seen that to conform to the basic figure of the 
method the gradients must be transformed into n in 1, m in 1 and 
k in 1. 



and 



h } = 



K = 



<*, 



k - n 



k - m 



w 



Side width W. = - + HB 
' 2 

- | + nh, 



(10.37) 
(10.38) 

(10.39) 



VOLUMES 519 



w 
W 2 = - + DJ 

w 

+ mh 2 (10.40) 



2 
Area of fill = Vih,d % 



*,' 



2(k - n) 
2(fc - n) 



(10.41) 



Area of cut = yih 2 d z 



•k- 



%k - m) 

2(fc - m) 

In the above ft has been treated as -ve, occurring in the 
cut. If it is +ve and the centre line is in fill, then 



(10.42) 



(? + **•)' 



Area of fill = K - (10.43) 

2(/c - n) 



/w \ 2 



Area of cut = (10.44) 

2(fc - m) 

N.B. If h = and m = «, 

w 2 

Area of cut = Area of fill = — (10.45) 

&(k - m) 

Example 10.4 A proposed road is to have a formation width of 40 feet 
with side slopes of 1 in 1 in cut and 1 in 2 in fill. The ground falls 
at 1 in 3 at right-angles to the centre line which has a reduced level 
of 260*3 ft. If the reduced level of the road is to be 262*8 ft, calcu- 
late (a) the side width, (b) the area of cut,(c) the area of fill. 

w w 

4 = 2 - <-M) = 2 + ok 

= 20 + 2*5 x 3 = 27*5 ft 



520 



SURVEYING PROBLEMS AND SOLUTIONS 




W, = 75-0" 



W 2 = 26-5' 

« 5 *. 



+ 262-8 
+260-3 







Fig. 10.20 


d 2 




= 20 - 7-5 


= 12-5 ft 


h. 


*i 


27-5 
"3-2 


= 27-5 ft 


k - n 


1u 


d z 


12-5 


= 6-25 ft 



k — m 



w 



W, = - + nh, = 20 + 2 x 27*5 = 75-0 ft 



W 2 = - + mh 2 = 20 + 1 x 6-5 = 26'5 ft 

Area of cut = l / 2 d 2 h 2 = % x 12*5 x 6*25 = 39-06 ft 2 
Area of fill = 1 / 2 d,ft, = Vi x 27*5 x 27-5 = 378*13 ft 2 

By Eqs. 10.43/10.44, 

.2 



Area of cut = 



Area of fill = 



G- tt °) 



2(k-m) 



(=♦«.)' 



2(k - n) 



(20 - 7-5) 
2(3 - 1) 



(20 + 7-5)' 
2(3 - 2) 



= 39-06 ft : 



= 378-13 ft 2 



Example 10.5 An access road to a small mine is to be constructed 
to rise at 1 in 20 across a hillside having a maximum slope of 1 in 10. 
The road is to have a formation width of 15 ft, and the volumes of cut 
and fill are to be equalised. Find the width of cutting, and the volume 
of excavation in 100 ft of road. Side slopes are to batter at 1 in 1 in 
cut and 1 in 2 in fill. 

(N.R.C.T.) 
To find the transverse slope (see page 413) 



VOLUMES 



521 




Fig. 10.21 



Let AB be the proposed road dipping at 1 in 20 (20 units) 
AC the full dip 1 in 10 (10 units) 
AD the transverse slope 1 in t (t units). 

In triangle ABC, 



COS0 = 



10 



In triangle ADC, 



= 11*55 (gradient value) 



Fig. 10.22 




If area of cut = area of fill, from Eqs. (10.43) and (10.44) for 



h +ve, 



(}-k„) 2 d + 4 



2(k - m) 



2{k - n) 



(7-5 - ll'SShf (7-5 + ll-SShf 



i.e. 



11-55 - 1 
7-5 - ll-55/i 



V 9. 



11-55 - 2 

55 



(7-5 + ll-55/i) 



9-55 

= 1-051(7-5 + ll-55fc) 
— 0*383 
H = 13^89 = - '° 1617 <*■«•*"*> 

x = kh = 11-55 x 0-01617 
= -0-187 ft 

Width of cutting = 7-5 + 0-187 = 7-687 say 7*69 ft 



522 



SURVEYING PROBLEMS AND SOLUTIONS 



Area of cutting 



(7-5 + 0-187) ! 
2(11-55 - 1) 
7-69 2 



21-10 
2-80 ft 2 



Volume of cutting = 2-80 x 100 ft 3 
= 10-37 yd 3 

(d) Sections with variable crossfall (three-level section) 

If the cross-section is very variable, it may be necessary to deter- 
mine the area either (a) by an ordinate method or (b) by plotting the 
section and obtaining the area by scaling or by planimeter. 

If the section changes ground slope at the centre line the follow- 
ing analysis can be applied, Fig. 10.23. 




Fig. 10.23 Section with variable crossfall 
The total area is made up of four parts: 
(1) Triangle AHB Area = i-h } d, 



(2) Trapezium BHGC Area = ^{h + fc, ) 

(3) Trapezium CGFD Area = j(h + h 2 ) 

(4) Triangle DFE Area = !fc 2 d 2 

Total Area = \[h,d,+ h 2 a\ + |(2fc + /i, + h 2 )] 



w 



Here fc, = h - - 



K = K + 



w 
27 



<*, = 



h x mk 

k + m 



(10.46) 



VOLUMES 



523 



, _ hzinl 

2 = 7^- 



Side width 



*» 



w 



+ 4 



w 



». - 2 + 4 

N.B. fc and / have both been assumed +ve, and the appropriate 
change in sign will be required if GA and GE are different from that 
shown. 

// the level of the surface is known, relative to the formation level, 
at the edges of the cutting or embankment (Fig. 10.24) 




Fig. 10.24 Section with levels at formation pegs 
Area ABCDHA = Area XBDH - Area XBA 



= J (f + <*,)(#, + h ) -H,d] 



Area DEFGH = Area DFYH - Area FYG 



Total Area 



- m(fi* + H 2 2 )j. 



,) 



(10.47) 



Exercises 10(b) (Cross-sectional areas) 

9. At a point A on the surface of ground dipping uniformly due South 
1 in 3, excavation is about to commence to form a short cutting for a 
branch railway bearing N 30° E and rising at 1 in 60 from A . The 



524 SURVEYING PROBLEMS AND SOLUTIONS 

width at formation level is 20 ft and the sides batter at 1 vertical to 
1 horizontal. 

Plot two cross-sections at points B and C 100 ft and 150 ft respec- 
tively from A and calculate the cross-sectional area at B. 

(N.R.C.T. Ans. 1323* 3 ft 2 ) 

10. Calculate the side widths and cross-sectional area of an em- 
bankment to a road with a formation width of 40 ft. The sides slope 
1 in 2 when the centre height is 10 ft and the existing ground has a 
crossfall of 1 in 12 at right-angles to the centre line of the embank- 
ment. 

(N.R.C.T. Ans. 34*28 ft; 48'01 ft; 622*8 ft 2 ) 

11. A road is to be constructed on the side of a hill having a cross- 
fall of 1 vertically to 8 horizontally at right-angles to the centre line 
of the road; the side slopes are to be similarly 1 to 2 in cut and 1 to 
3 in fill; the formation is 50 ft wide and level. Find the distance of 
the centre line of the road from the point of intersection of the forma- 
tion with the natural ground to give equality of cut and fill, ignoring 
any consideration of 'bulking'. 

(L.U. Ans. 1*14 ft on the fill side) 

12. A road is to be constructed on the side of a hill having a cross- 
fall of 1 vertically to 10 horizontally at right-angles to the centre line 
of the road; the side slopes are to be similarly 1 to 2 in cut and 1 to 

3 in fill; the formation is 80ft wide and level. Find the position of 
the centre line of the road with respect to the point of intersection of 
the formation and the natural ground, (a) to give equality of cut and 
fill, (b) so that the area of cut shall be 0*8 of the area of fill in order 
to allow for bulking. 
(L.U. Ans. (a) 1-34 ft on the fill side; (b) 0*90 ft on the cut side) 

13. The earth embankment for a new road is to have a top width of 
40 ft and side slopes of 1 vertically to 2 horizontally, the reduced 
level of the top surface being 100*0 O.D. 

At a certain cross-section, the chainages and reduced levels of 
the natural ground are as follows, the chainage of the centre line being 
zero, those on the left and right being treated as negative and positive 
respectively: 

Chainage (ft) -50 -30 -15 -0 +10 +44 
Reduced level (ft) 86*6 88*6 89*2 90*0 90*7 92*4 
Find the area of the cross- section of the filling to the nearest 
square foot, by calculation. 

(L.U. Ans. 567 ft 2 ) 

14. A 100 ft length of earthwork volume for a proposed road has a 
constant cross-section of cut and fill, in which the cut area equals the 



VOLUMES 



525 



fill area. The level formation is 30 ft wide, the transverse ground 
slope is 20° and the side slopes in cut and fill are respectively 
^(horizontal) to 1 (vertical) and 1 (horizontal) to 1 (vertical). 
Calculate the volume of excavation in 100 ft length. 

(L.U. Ans. 209-2 yd 3 ) 

10.32 Alternative formulae for the calculation of volumes from the 
derived cross-sectional areas 

Having computed the areas of the cross-sections, the volumes in- 
volved in the construction can be computed by using one of the ordinate 
formulae but substituting the area of the cross-section for the ordinate. 

(1) Mean Area Rule 

W 
v = -04, +A 2+ A a + ... +AJ 



i.e. 



n 



(10.48) 



where W = total length between end sections measured along centre 
line. 
n = no. of sectional areas. 
2-4 = sum of the sectional areas. 
N.B. This is not a very accurate method. 

(2) Trapezoidal (or End Area) Rule (Fig. 10.25) 





< 

A 2 


"3 
*3 




>.'• 


r^ 


A 


- * . 


- *2 . 


. W* - 


* w * » 


- "* » 


. ** . 




. w ~ 


iw 









Fig. 10.25 Trapezoidal rule 

y z = T 2 (a z + a 3 ) 

w 
V 3 = ^(A 3+ A 4 ) 

2 



■&4n-l + K) 



526 



SURVEYING PROBLEMS AND SOLUTIONS 



If W, = W 2 = W n , then 

V = (V, + V 2 + V 3 + ... +V n _,) 
w 
2 



= - U, + 2A 2 + 2A 3 + 2A A + ... +2A n _, + 4J 

(10.49) 



(3) Prismoidal Rule (Fig. 10.26) 

As the cross-sections are all parallel and the distance apart can 
be made equal, the alternate sections can be considered as the mid- 
section. 

The formula assumes that the mid-section is derived from the mean 
of all the linear dimensions of the end areas. This is difficult to apply 
in practice but the above application is considered justified particularly 
if the distance apart of the sections is kept small. 



'1 




"3 
*3 


^4 
*4 






M W - 


. w » 

















Fig. 10.26 Prismoidal rale 

Thus (V, + V 2 ) = ^04, + 4A 2 + A 3 ) 


(V, + V 4 ) = |(^ 3 + 4A 4 + A s ) 

(V 8 + V 6 ) = |o4 5 + 4A 6 + A 7 ) 

w 
.% Total volume = -U, + 4A 2 + 2A 3 + 4A A + 2A S + 4A 6 + A 7 ] 

If the number of sections is odd, then 

w 
V = -3 [4, + 41 even areas + 22 odd areas + 4J (10.50) 

which is Simpson's rule applied to volumes. 

Prismoidal Corrections 

If having applied the end areas rule it is then required to find a 
closer approximation, a correction can be applied to change the deri- 
ved value into the amount that would have been derived had the pris- 
moidal rule been applied. For areas /I, and A 2 s units apart, 

By the end areas formula, V^, = —(,4, + A 2 ) 



VOLUMES 527 

s, 



By the prismoidal formula, V P = t(^i + 4A m + A 2 ) 

The difference will be the value of the correction, i.e. 

= v e - V P = |[3A, + 3A 2 - A, - AA m - A 2 ] 



c 



6 



s 



c = H204, + A 2 ) - 4A m ] (10.51) 



6 



For sections without crossfall 

Let the two end sections A y and A 2 be s it apart with formation 
width of w ft and formation heights h r and h 2 . 

Then, by Eq. (10.26), 

A^ = A, (w + mh y ) 
A 2 = h 2 (w + mh 2 ) 

Putting these values into Eq. (10.51), 

c = I^IwCft, + h 2 ) + m(h* + h 2 )\ 

- 2\wQi, + h 2 ) + f (/if + hi + 2h,h 2 )\] 
= ™[2h? + 2hl - h z - hi - 2h<h 2 ] 

c = ~?U, - fc 2 ) 2 (10.52) 

6 

For sections with crossfall 
From Eq. (10.34), 

m\h z k z +— w z + wh.ml 

A % = — LJ _±__ !_ J + w /i, 

(fc 2 -m 2 ) 

m\h z k z + -w 2 + w/i,/m| 

4, = — ^ 1 3 — J + wfc 2 

2 (k 2 - m 2 ) 

m[i(/i 1 + fc 2 ) 2 k 2 + ^w 2 + >(ft, + h 2 )] , 

^4 m = — — r- 5 — 5 +-w(h.+h 9 ) 

(fc 2 -ro 2 ) 2 ' 2 

Substituting these values in Eq. (10.51), 

Prismoidal Correction c = ^[2C4, + A 2 ) - 4A m ] 



528 SURVEYING PROBLEMS AND SOLUTIONS 

c = ar^ zP * fc W + n t ) + I w 2 + wm(h, + h z ) 
oC/c 4 - m ) 

+ wm(k 2 - m z )(h x + h 2 ) \ 
-4{±k 2 (ht + h 2 ) 2 + {w* 
+ ±wm(k 2 -m 2 )(h x +h 2 )}] 

c = 6( fe/ m m2) [2Jc 2 (ftf + fc 2 2 ) - /c 2 (/i, + fc 2 ) 2 ] 

sm/c (ft, - /i 2 ) 2 
C 6(fc 2 - m 2 ) 

For sections with cut and fill 
From Eq. (10.42), (^ + kh\ 



(10.53) 



A. 



K 



-rim — 



2(fc - m) 

(I + kh *) Z 

2(k - m) 



2(k - m) 
Substituting these values in Eq. (10.51), 

Prismoidal correction = f \ 2 l(™ + kh \ 2 + (l +kh \ 2 ) 

for cut 12(k-m)L \\2 7 V 2 7 J 

-«{H». ♦*.>)'] 

Prismoidal correction _ s [? / / * _ kh¥ + (- - kh \ 2 1 

for fill = 12(fc -n)[ l\2 / \2 7 / 

= s/c 2 ^ - ft 2 ) 2 Q 

12 (k - n) 

Example 10.6 An embankment is to be formed with its centre line on 
the surface (in the form of a plane) on full dip of 1 in 20. If the 
formation width is 40 ft and formation height are 10, 15, and 20 ft 
at intervals of 100 feet, with the side slopes 1 in 2, calculate the 
volume between the end sections. 



VOLUMES 



529 




Fig. 10.27 Longitudinal section 



40' 



^ Ay 



ho' 



|£1< j^ A, 20' 



Fig. 10.28 Cross-sections 

Area (1) = h y (yv + mh } ) 

= 10(40 + 2 x 10) 

= 600 ft 2 

Area (2) - 15(40 + 2 x 15) 

= 1050 ft 2 

Area (3) = 20(40 + 2 x 20) 

= 1600 ft 2 



Volume 

(1) By Mean Areas 



W 
V = £<2i4) 



200 



(600 + 1050 + 1600) 



= 216 666-7 ft c 



530 SURVEYING PROBLEMS AND SOLUTIONS 



(2) By End Areas (Trapezoidal) 

V = ^[A t + 2A 2 + A 3 ] 

= ^[600 + 2100 + 1600] 
= 215000 ft 3 



(3) By the prismoidal rule (treating the whole as one prismoid) 
V = |U, + 4A 2 + A 3 ] 

= ™ [600 + 4200 + 1600] 
= 213 333-3 ft 3 



(4) By Prismoidal Correction to End Areas 

(By end Areas _ 100 + _ 82m ft3 

in each section) I 

V 2 = ^(1050 + 1500) = 132500 ft 3 



= 215000 ft 2 



By outer Areas V = -^(600 + 1600) = 220000 ft 3 



Applying Prismoidal Correction to adjacent areas, 

<y E - v p\ = ^pOh -fc.) a 

= 10 ° x 2 (i 5 __ 10) 2 = 833-33 ft 3 
o 

(V E - V P ) 2 = 10 ° 6 X 2 (20 - 15) 2 = 833-33 ft 3 

Total Correction = + 1666-66 ft 3 
.". Volume V p =215000 - 1666-67 = 213 333-33 ft 3 

Applying Prismoidal Correction to outer areas, 



V e ~ V p = 2 °° 6 X 2 (20 - 10) 2 = 6666-67 

V P = 220000-6666-67 = 213 333-33 ft 5 



VOLUMES 



531 



N.B. (1) The correct value is only obtained by applying the prismoidal 
correction to volumes obtained by adjacent areas unless, as here, the 
whole figure is symmetrical. 

(2) The prismoidal correction has little to commend it in prefer- 
ence to the application of the prismoidal formula if all the information 
is readily available. 

Example 10.7 Given the previous example but with the centre line 
turned through 90° 



40' 




Fig. 10.29 Cross-sections 



From Eq. (10.34), 



m[h 2 k z + -w 2 + whm] 

A = + wh 

kr - m z 



Cross-sectional Areas 
2 



, - 20^72^ [1 ° 2 x 20 2 + J- x 40 2 + 40 x 10 x 2] + 40 x 10 



A. = 



198 

J. 
198' 

1 



[40000 + 400 + 800] + 400 = 608*08 ft 2 



K = ^[90000 + 400 + 1200] + 600 = 1062-62 ft 5 

A„ = 



198 



[160000 + 400 + 1600] + 800 = 1618- 18 ft 2 



Volume 

(1) By Mean Areas (Eq. 10.48) 

V = ^[608-08 + 1062-62 + 1618-18] = 219258-7 ft 3 



(2) By End Areas (Eq. 10.49) 

(taking all y = 100 [60g . 08 106 2-62 + 1618-18] - 217575 ft 3 
sections) 2 

(taking outer y _ 2W [60g . 08 + „„.„, _ 222626 ft , 
sections) 2 



532 



SURVEYING PROBLEMS AND SOLUTIONS 



(3) By the prismoidal rule (Eq. 10.50) (treating the whole as one 
prismoid) 

V = ^ [608-08 + 4 x 1062-62 + 1618-18] = 215891-5 ft 3 

(4) By applying Prismoidal Correction to End Areas 
Applying prismoidal correction to each section (Eq. 10.53), 

Smk 2 



c = 
c 2 = 



6(/r - to 2 ) 

100 x 2 x 20 2 
6 x 396 

100 x 2 x 20 2 



396 



2^ x (ft, - h 2 ) 

x (15 - 10) 2 = 841-75 ft* 
x (20 - 15) 2 = 841-75 iV 



= 1683-50 

.-. V P = 217 575 - 1683-5 = 215 891-5 ft 5 

Applying prismoidal correction to outer areas, 

c . **.* 2 .?. 202 x (20 - 10) 2 . 6734 ft- 



6 x 396 
V P = 222626 - 6734 



= 215892 ft 3 



N.B. Where the figure is symmetrical the prismoidal correction again 
gives the same value, but it would be unwise to apply the latter method 
where the prismoids are long and the cross-sectional areas very vari- 
able, unless it is applied to each section in turn, as shown above. 

Example 10.8 A road has a formation width of 40 ft, and the side 
slopes are 1 in 1 in cut and 1 in 2 in fill. The ground slopes at 1 in 
3 at right-angles to the centre line. Sections at 100 ft centres are 
found to have formation heights of + 1 ft, 0, and -2 ft respectively. 
Calculate the volumes of cut and fill over this length. 



Fig. 10.30 




VOLUMES 533 



Areas of cut 

From Eq. (10.44), 



(! - kh Y 

(h + ve) Area of Cut = x - 7f71 ^ 

2(/c - m) 



(f - 3_x 1) 

(20 - 0) 2 



A < ' 2(3 - 1) = 72 ' 25 ft2 



A 2 = ^ — '— = 100 ft 2 



4 

.2 



(ft _ ve) . - (2 ° t 3 X 2y - 169 ft' 



Areas o/ /i/Z 

From Eq. (10.43), 



(h + ve) Area of fill 



••• 


K = 


(20 + 3) 
2(3 - 2) 




A' z = 


(20 + 0) 2 
2 


(/z-ve) 


K = 


(20 - 6) 2 
2 


Volume of Cut 
(1) By Mean i4 


reas 





2(fc --n) 
= 264-5 ft 2 

= 200 ft 2 



98 ft : 



V = 2 25[ 72 -25 + 100 + 169] = 22750 ft 3 



(2) By End Areas 

(taking all sections) 



V = l°-P-[72-25 + 2 x 100 + 169] = 22062-5 ft 3 
(taking outer sections) 

V = ^[72-25 + 169] = 24 125 ft 3 

(3) By the prismoidal rule (treating the whole as a prismoid) 

V = ^[72-25 + 4 x 100 + 169] = 21375 ft 3 



534 SURVEYING PROBLEMS AND SOLUTIONS 

(4) By applying Prismoidal Correction to End Areas 

Applying prismoidal correction to each section of cut, 

c = Sk\h, - h 2 ) 2 
12(k - m) 

... c = 100 x 3 2 (1 - 0) 2 _ 37 . 5 , 

1 12(3 - 1) ~ 6/ * " 

m 900 x (2 - Of m ft3 

2 24 

c T = 187-5 ft 3 

.-. V P = 22062-5- 187-5 = 21875-0 ft 3 



Applying prismoidal correction to outer areas, 

900 (1 + 2) 2 „ n r _ 

c = L '— = 337.5 ft 3 

24 
.-. V P = 24125 - 337*5 = 23787-5 ft 3 

Volumes of Fill 

(1) By Mean Areas 

V = -^[264-5 + 200 + 98] = 37500 ft 3 

(2) By End Areas 

(taking all sections) 

V = ^[264-5 + 2 x 200 + 98] = 38 125 ft 3 



(taking outer sections) 

V = ^[264-5 + 98] = 36250 ft 3 

(3) By the prismoidal rule (treating the whole as a prismoid) 

V = y^264-5 + 4 x 200 + 98] = 38750 ft 3 

(4) By applying Prismoidal Correction to End Areas 
Applying prismoidal correction to each section of fill, 

100 x 3 2 x l 2 



C ' 12(3 - 2) 


= 75 ft 3 


900 x 2 2 
C > = 12 


= 300 ft 3 


c T 


= 375 ft 3 


V P = 38 125 - 375 


= 37750 fta 



VOLUMES 535 



Applying prismoidal correction to outer areas, 
c = 



Vp = 36250 - 675 = 35 575 ft 3 

N.B. The prismoidal correction applied to each section gives a much 
closer approximation to the value derived by the prismoidal formula, 
although the latter in this case is not strictly correct as the middle 
height is not the mean of the two end heights. 

Example 10.9 Calculate the volume between three sections of a rail- 
way cutting. The formation width is 20 ft; the sections are 100 ft 
apart; the side slopes are 1 in 2 and the heights of the surface above 
the formation level are as follows: 

Section Left Centre Right 

1 17-6 16-4 17-0 

2 21-2 20-0 18-8 

3 19-3 17-9 16-3 
From Eq. (10.47), 



Area = 1 |f + nitty (H, + h ) + (| + mfty(H 2 + h ) 

- m(H? + H 2 * )j 
Section 1, A, = ± U y + 2 x 17-e) (17-6 + 16*4) 

+fy + 2x 17-o)(17-0 + 16-4) 
- 2(17-6 2 +17-0 2 ) = 904-4 ft 2 

Section 2, A 2 = I [(10+ 42-4) (41*2) + (10 + 37-6) (38-8) 
- 2(21-2 2 + 18-8 2 )] = 1200-0 ft 2 

Section 3, A 3 = j[(10 + 38-6)(37-2) + (10 + 32-6)(34-2) 
+ 2(19-3 2 +16-3 2 )] = 994-2 ft 2 
Using the prismoidal formula, 

V = 3^7t904-4 + 4x 1200 + 994-2] = 8270 yd 3 

Total Volume = 8 270 yd 3 

10.33 Curvature Correction (Fig. 10.31) 

When the centre line of the construction is curved, the cross- 
sectional areas will be no longer parallel but radial to the curve. 



536 



SURVEYING PROBLEMS AND SOLUTIONS 



Volume of such form is obtained by using the Theorem of Pappus 
which states that 'a volume swept out by a constant area revolving 
about a fixed axis is given by the product of the area and the length of 
the path of the centroid of the area*. 

The volume of earthworks involved in cuttings and embankments 
as part of transport systems following circular curves may thus be 
determined by considering cross-sectional areas revolving about the 
centre of such circular curves. 



Area A, 



Centroid 



Section at X 



Centre line 
of formation 



Centre of 
curvature 



& 



Centroid 




Section at Y 



Fig. 10.31 Curvature correction 

If the cross-sectional area is constant, then the volume will equal 
the product of this area and the length of the arc traced by the centroid. 

If the sections are not uniform, an approximate volume can be de- 
rived by considering a mean eccentric distance (e) = e * + ez relative 

to the centre line of the formation. 

This will give a mean radius for the path of the centroid (/? ± e), 
the negative sign being taken as on the same side as the centre of 
curvature. 

Length of path of centroid XY = (R + e) rad . 



but 



&ra.d = "o where S = length of arc on the centre 

line 



.-. XY = |(K + e) = s(l ± |) 



VOLUMES 



537 



Volume is given approximately as 

v = |(A + 4,)(i±f) 



(10.56) 



Alternatively each area may be corrected for the eccentricity of 
its centroid. 

If e, be the eccentricity of the centroid of an area A, , then the 
volume swept out through a small arc 8d is SV = A^(R ± e,)50. 
If the eccentricity had been neglected then 

8V = A.R80 
with a resulting error = A^e^bd 

_ Vi 



R 



per unit length 



(10.57) 



Ae 



Thus, if each area is corrected by an amount ±~H"> these new 
equivalent areas can be used in the volume formula adopted. 

10.34 Derivation of the eccentricity e of the centroid G 

Centroids of simple shapes 
Parallelogram (Fig. 10.32) 

G lies on the intersection of the diagonals or the intersection of 
lines joining the midpoints of their opposite sides. (10.58) 




Fig. 10.32 




Triangle (Fig. 10.33) 
G lies at t 

from each apex 



G lies at the intersection of the medians and is -=- of their length 

(10.59) 



538 SURVEYING PROBLEMS AND SOLUTIONS 

Trapezium (Fig. 10.34) 






(10.60) 




Fig. 10.35 



Fig. 10.34 

A Compound Body (Fig. 10.35) 

If the areas of the separate parts are A x and A z and their cen 
troids G, and G z , with the compounded centroid G, 

G = A 2 xG,G 2 
1 A, + A z 

A, x G,G 2 



or G 2 G = 



^i + A z 



(10.61) 
(10.62) 



Thus for typical cross-sectional areas met with in earthwork cal- 
culations, the figures can be divided into triangles and the centre of 
gravity derived from the compounding of the separate centroids of the 
triangles or trapezium, Fig. 10.36. 




Fig. 10.36 
Alternatively, Fig. 10.37, 

Let the diagonals of ABCD intersect at E. 

BO = OD on line BD 
AE = FC on line AC 
then 2.0G = GF 



Fig. 10.37 



(10.63) 



VOLUMES 



539 



To find the eccentricity e of the centroid G 

Case 1. Where the surface has no crossfall, the area is symmetrical 

and the centroid lies on the centre line, i.e. e - 0. 

Case 2. Where the surface has a crossfall 1 in k (Fig. 10.38) 

Let Total Area of ABDE = A T 

Area of triangle AEF 




*-«, 



= \AF(H 2 -H,) 
_*W+ W 2 ) 



(10.64) 



B I . wlz 7 F ° 

Fig. 10.38 Section with crossfall 
Let G, and G 2 be the centroids of areas AEF and AFDB respectively. 
Length AQ = horizontal projection of AG y 

m I pi? + AFl = i^ + ^ +2R;] 



= iUW, + WJ = W, + -^ 



Distance of Q from centre line, 



W, 



i.e. XQ, = G 2 g, - HJ +J18-- Hf, 



(10.65) 



Distance of centroid G for the whole figure (from the centre line, i.e. 
e), 

Area A 4EF x XQ ¥f y W Z (W, + W 2 ) 



e = 



Total Area A, 



3k. A, 



(10.66) 



Conversion Area A c - ± 



A r e 
R 



540 



SURVEYING PROBLEMS AND SOLUTIONS 



i.e. 



A = + A T [H\W 2 (W, + W 2 )] 
W,W 2 (W, + W 2 ) 



Corrected Area = A T ± 



3k R 

W,W 2 (W, + W 2 ) 
ZkR 



Case 3. Sections with part cut and part fill (Fig. 10.39) 



(10.67) 



< 


^7 




J 

e2 > 




f '« 


^ 


^ 2 



Fig. 10.39 Section part cut/part fill 



For section in cut, i.e. triangle CED, G lies on the median EQ. 
J Q = I (f + *) ~ x = I ("f - *) 



e 2 = JQ + i(W 2 - JQ) = I(^ + 2JQ) 



-*(« 



w 



W 2 + j- *ft . 



) 



Similarly for fill e, = j (W, + ^ + &O 



(10.68) 
(10.69) 



Example 10.10 Using the information in Example 10.6, viz. embank- 
ment with a surface crossfall of 1 in 20, side slopes 1 in 2, formation 
width 40 ft and formation heights of 10, 15 and 20 ft at 100 ft cent- 
res, if this formation lies with its centre line on the arc of a circle of 
radius 500 ft, calculate 

(a) the side widths of each section, 

(b) the eccentricity of their centroids, 

(c) the volume of the embankment over this length for the centre 
of curvature (i) uphill (ii) downhill. 



VOLUMES 



541 




Fig. 10.40 



(a) Side widths 
Section 1 

From Eq. (10.31), 



W, = 



(*• " I) 



mk 



w 

2 k + m 



and from Eq. (10.32), 



W„ = 



(*° - fk) mk 



w 

2 k - m 



40 
i.e. W f = -V + 



*» " D 



2 x 20 



2 ' 20+2 

9 x 40 



Section 2 



Section 3 



W, = 



w. = 



w„ = 



w. = 



w„ = 



20 + 
20 + 

20 + 

20 + 

20 + 
20 + 



22 
11 x 40 
18 



14 x 40 
22 

16 x 40 
18 



19 x 40 
22 

21 x 40 
18 



= 36-36 ft 





44-44 ft 




45-45 ft 




55-56 ft 




54-55 ft 


_ 


66-67 ft 



(b) Eccentricity (e) 
From Eq. (10.66), e 



3k A 



542 SURVEYING PROBLEMS AND SOLUTIONS 



_ 36-36 x 44-44 (36-36 + 44-44) 
?1 3 x 20 x 608-08 



= 3-58 ft (Area 608*08 ft 2 from previous calcula- 
tions) 

= 45-45 x 55-56 (45-45 + 55-56) 
&z 3 x 20 x 1062-62 

= 4-00 ft 

= 54-55 x 66-67 (54-55 x 66-67) 
63 3 x 20 x 1618-18 

= 4-54 ft 

(c) Volumes 

Using the above values of eccentricity in the prismoidal formula, 
the volume correction 

V . ± M[ 608 -08 x |f ♦ 4 (1062-62 , ^) + 1618-18 x±|] 

= ± ^[608-08 x 3-58 + 16 x 1062-62 + 1618-18 x 4-54] 
= ±1768-4 ft 3 



The correction is + ve if the centre of the curve lies on the uphill 
side. 

.*. Corrected volume = 215891-5 ± 1768*4 ft 3 

= 217 660 ft 3 

or 214 123 ft 3 



A more convenient calculation of volume, without separately cal- 
culating the eccentricity, is to correct the areas using Eq. (10.67). 

Area Correction A„ = ± • 2V ! 11. 

Z.k.R 

A - + 36 ' 36 x 44 ' 44 (36 ' 36 + 44 ' 44) 
°' " " 3 x 20 x 500 



= ± 4-35 ft 



A = ± 45 ' 45 x 55 ' 56 ( 45 ' 45 + 55 ' 56 ) 
° 2 3 x 20 x 500 

= ± 8-50 ft 2 

A = + 54-55 x 66-67 (54-55 + 66-67) 
° 3 3 x 20 x 500 

= ± 14-70 ft 2 



VOLUMES 543 



Corrected Areas are: 

A, = 608-08 ±4-35 = 612-43 

or 603-73 ft 2 

A 2 = 1062-62 ± 8-50 = 1071-12 

or 1054-12 ft 2 

A 3 = 1618-18 ±14-70 = 1632-88 

or 1603-48 ft 2 

Corrected volumes: 

(i) With centre of curve on uphill side, 

V - ^[612-43 + 4 x 1071-12 + 1632-88] 
- 217657 ft 3 



(ii) With centre of curve on downhill side, 

V = ^[603-73 + 4 x 1054-12 + 1603-48] 
= 214122 ft 3 



10.4 Calculation of Volumes from Contour Maps 

Here the volume is derived from the areas contained in the plane 
of the contour. For accurate determinations the contour interval must 
be kept to a minimum and this value will be the width (w) in the form- 
ulae previously discussed. 

The areas will generally be obtained by means of a planimeter, 
the latter tracing out the enclosing line of the contour. 

For most practical purposes the Prismoidal formula is satisfactory, 
with alternate areas as 'mid-areas' or, if the contour interval is large, 
on interpolated mid-contour giving the required 'mid-area' may be used. 

10.5 Calculation of Volumes from Spot-heights 

This method uses grid levels from which the depth of construction 
is derived. 

The volume is computed from the mean depth of construction in 
each section forming a truncated prism, the end area of which may be 
rectangular but preferably triangular, Fig. 10.41. 

V = plan area x mean height (10.70) 

If a grid is used, the triangular prisms are formed by drawing 
diagonals, and then each prism is considered in turn. 



544 



SURVEYING PROBLEMS AND SOLUTIONS 




/>, t 


2 h 


\ 


/ 




/ 


\ 






/ 




/ 




s 




t 








/ 


"4 \ 


,'th 


• V 










/ 




/ 


























\ 








N 



/»• 



"7 



Fig. 10.41 Volume from spot-heights 



Fig. 10.42 



The total volume is then derived (each triangle is of the same 
area) as one third of the area of the triangle multiplied by the sum of 
each height in turn multiplied by the number of applications of that 
height, 



i.e. V = ^[Inh] 



(10.71) 



e.g. V = -[2/i, + 2hz + 2h 3 + 2h 4 + 8h 5 + 2h 6 + 2hy + 2h e + 2h 9 ] 



(Fig. 10.42) 



10.6 Mass-haul Diagrams 



These are used in planning the haulage of large volumes of earth- 
work for construction works in railway and trunk road projects. 

10.61 Definitions 

Bulking An increase in volume of earthwork after excavation. 
Shrinkage A decrease in volume of earthwork after deposition and 
compaction. 

Haul Distance (d) The distance from the working face of the excava- 
tion to the tipping point. 

Average Haul Distance (D) The distance from the centre of gravity 
of the cutting to that of the filling. 

Free Haul Distance The distance, given in the Bill of Quantities, 
included in the price of excavation per cubic yard. 
Overhaul Distance The extra distance of transport of earthwork 
volumes beyond the Free Haul Distance. 

Haul The sum of the product of each load by its haul distance. This 
must equal the total volume of excavation multiplied by the average 
haul distance, i.e. S.v.d = V.D. 



VOLUMES 



545 



Overhaul The products of volumes by their respective overhaul dis- 
tance. Excess payment will depend upon overhaul. 
Station Yard A unit of overhaul, viz. 1 yd 3 x 100 ft. 
Borrow The volume of material brought into a section due to a 
deficiency. 
Waste The volume of material taken from a section due to excess. 

In S.I. units the haul will be in m 3 , the haul distances in metres 
and the new 'station' unit probably 1 m 3 moved 100 m. 

10.62 Construction of the mass-haul diagram (Fig. 10.43) 

(1) Calculate the cross- sectional areas at given intervals along 
the project. 

(2) Calculate the volumes of cut and fill between the given areas 
relative to the proposed formation. 

N.B. (a) Volumes of cut are considered positive, 
(b) Volumes of fill are considered negative. 

(3) Calculate the aggregated algebraic volume for each section. 

(4) Plot the profile of the existing ground and the formation. 

(5) Using the same scale for the horizontal base line, plot the 
mass haul curve with the aggregated volumes as ordinates. 



M P 



Max. haulage 
point 

Mass haul diagram 

Fig. 10.43 Mass-haul curves 




546 SURVEYING PROBLEMS AND SOLUTIONS 

10.63 Characteristics of the mass-haul diagram 

(1) A rising curve indicates cutting as the aggregate volume is in- 
creasing (a—/ is seen to agree with AF on the profile). 

(2) A maximum point on the curve agrees with the end of the cut, 
i.e. f-F. 

(3) A falling curve indicates filling as the aggregate volume is 
decreasing (f-k is seen to agree with F-K on the profile). 

(4) The vertical difference between a maximum point and the next 
minimum point represents the volume of the embankment, i.e. //, + k,k 
(the vertical difference between any two points not having a minimum 
or maximum between them represents the volume of earthwork between 
them.) 

(5) If any horizontal line is drawn cutting the mass-haul curve (e.g. 
aqp), the volume of cut equals the volume of fill between these points. 
In each case the algebraic sum of the quantities must equal zero. 

(6) When the horizontal balancing line cuts the curve, the area 
above the line indicates that the earthwork volume must be moved for- 
ward. When the area cut off lies below the balancing line, then the 
earthwork must be moved backwards. 

(7) The length of the balancing line between intersection points, 
e.g. aq, qp, represents the maximum haul distance in that section (q 
is the maximum haulage point both forward, aq, and backwards, pq). 

(8) The area cut off by the balancing line represents the haul in 
that section. N.B. As the vertical and horizontal scales are different, 
i.e. 1 in. = s ft horizontally and 1 in. = v yd 3 , an area of a in 2 repre- 
sents a haul of avs yd 3 , ft = ^r-r station yards. 

10.64 Free-haul and overhaul (Fig. 10.44) 

The Mass-haul diagram is used for finding the overhaul charge as 
follows: 

Free-haul distance is marked off parallel to the balance line on 
any haul area, e.g. bd. The ordinate cc 2 represents the volume dealt 
with as illustrated in the profile. 

Any cut within the section ABB^ 4, has to be transported through 
the free- haul length to be deposited in the section D^E^ED. This rep- 
resents the 'overhaul' of volume (ordinate bb,) which is moved from the 
centroid G, of the cut to the centroid G 2 of the fill. 

The overhaul distance is given as the distance between the cen- 
troids less the free-haul distance. 

i.e. (G t G 2 ) - bd 



VOLUMES 



547 



Balance line 




(b) Profile 
Fig. 10.44 Free-haul and overhaul 



The amount of overhaul is given as the volume (ordinate bb % = dd,) 
x the overhaul distance. 

Where long haulage distances are involved, it may be more eco- 
nomical to waste material from the excavation and to borrow from a 
location within the free-haul limit. 

If / is the overhaul distance, c the cost of overhaul and e the 
cost of excavation, then to move 1 yd 3 from cut to fill the cost is 
given as 

e + Ic 

whereas the cost to cut, waste the material, borrow and tip without 
overhaul will equal 2e. 

Economically e + Ic = 2e 

I = — (assuming no cost for wasting) 

Thus if the cost of excavation is 2/6 per yd 3 and the cost of over- 
haul is 2d per station yard, then the total economic overhaul distance 

= -y = 1500 ft 



If the free-haul is given as 500 ft the maximum economic haul 
= 1500 + 500 = 2000 ft. 

The overhaul distance is found from the mass-haul diagram by de- 
termining the distance from the centroid of the mass of the excavation 
to the centroid of the mass of the embankment. 

The centroid of the excavation and of the embankment can be 



548 SURVEYING PROBLEMS AND SOLUTIONS 

determined (1) graphically, (2) by taking moments, (3) planimetrically. 
These methods are illustrated in the following example. 

Example 10.11 Volumes of cut and fill along a length of proposed 
road are as follows: 





Chainage 




Volume 
Cut 


(ft 3 ) 
Fill 





100 




290 






200 




760 






300 




1680 






400 




620 






480 




120 






500 






20 




600 






110 




700 






350 




800 






600 




900 






780 




1000 






690 




1100 






400 




1200 






120 


Draw a mass 


diagram, and excluding the i 


surplus excavated mat- 


erial along this length determine the overhaul if the free-haul distance 


is 300 ft. 








(I.C.E.) 


Answer 










Chainage 


Volume (ft 3 ) A , 
Cut Fill Aggregate volume (ft 3 ) 



100 


290 






+ 290 


JL\J\J 

200 
300 
400 


760 






+ 1050 


1680 






+ 2730 


620 






+ 3350 


7UW 

480 


120 






+ 3470 


500 
600 




20 




+ 3450 




110 




+ 3340 


700 




350 




+ 2990 


/ \J\J 

800 
900 




600 




+ 2390 




780 




+ 1610 


1000 




690 




+ 920 


1100 




400 




+ 520 


1200 


3470 
3070 


120 
3070 




+ 400 


Check 


400 









VOLUMES 



549 



(ft 3 ) + 

4 OCX) 



3000 



2000 



1000 




100 200f 300 400 500 600 700 800 |900 1000 1100 1200ft 
240' 880' 

Fig. 10.45 Mass-haul diagram 



(1) Graphical Method 

(i) As the surplus of 400 ft 3 is to be neglected, the balancing 
line is drawn from the end of the mass-haul curve, parallel to the base 
line, to form a new balancing line ab. 

(ii) As the free-haul distance is 300 ft, this is drawn as a balanc- 
ing line cd. 

(iii) From c and d, draw ordinates cutting the new base line at 

(iv) To find the overhaul: 

(a) Bisect cc, to give c 2 and draw a line through c 2 parall- 
el to the base line and cutting the curve at e and /, which 
now represent the centroids of the masses acc % and dbd, 

(b) The average haul distance from acc^ in excavation to make 
up the embankment dbd x = ef. 

(c) The overhaul distance = the haul distance-the free-haul 
distance, i.e. 

ef - cd 

i.e. Scaled value = 640 - 300 = 340 ft. 



(d) The overhaul of material at acc x 

= volume (cc,) x overhaul distance (ef 
= 2750 ft 3 x 340 ft 
= 346' 3 station yards 



cd) 



(2) By taking moments 

With reference to the mass-haul curve and the tabulated volumes, 
moments are taken at a to find the centroid of the area acc^ . 

At a the chainage is scaled as 120 ft. 



550 



SURVEYING PROBLEMS AND SOLUTIONS 



Chainage 
(a) 120 - 200 

200 - 300 
300 - 350 (c) 



Volume (ft 3 ) 
1050-400 = 650 

2730- 1050 = 1680 



3150-2730= 420 

Iv = 2750 
Thus the distance from a to the centroid 
330500 



Distance (ft) 
|(200-120) = 40 

^(300-200) + 80 = 130 

| (350 -300)+ 180 = 205 



Product (VxD) 

26000 

218400 

86100 

» 330 500 



2750 



= 120-2 ft 



/. Chainage of the centroid = 120 + 120-2 = 240*2 ft. 
Taking moments at d, chainage 650 ft: 



Chainage 


Volume (ft 3 ) 


Distance (ft) 


Product (VxD) 


(d) 650 - 700 


3350-2990 = 160 


J( 700- 650) = 25 


4000 


700- 800 


2990-2390 = 600 


|(800-700)+ 50 = 100 


60 000 


800- 900 


2390-1610= 780 


l(900-800) + 150 = 200 


156 000 


900-1000 


1610- 920 = 690 


|(1000 -900) + 250 = 300 


207 000 


1000-1100 


920- 520 = 400 


|(1 100 - 1000) + 350 =400 


160000 


1100-1200 - 


520- 400 = 120 


^(1200- 1100) + 450 = 500 


60 000 




2v =2750 




SP 647000 


Thus the dists 


Mice from d to the < 


:entroid 






647000 


= 235* 3 ft 





2750 

Chainage of the centroid = 650 + 235*3 = 885*3 ft 

Average haul distance = 885*3 - 240*2 = 645*1 

Length of overhaul = 645*1 - 300 = 345*1 

rt . f 2750 x 345*1 oc1 c . .. 

Overhaul = — -— = 351*5 station 



2700 



yards. 



(3) Planimetric Method 

Distance to centroid = Haul/volume 

_ Area x horizontal scale x vertical scale 
volume ordinate 
From area acc^ 

Area scaled from mass-haul curve = 0*9375 in 2 

Horizontal scale = 200 ft to 1 in. 



VOLUMES 



551 



Vertical scale = 1600 ft to 1 in. 
.-. Haul = 0-9375 x 200 x 1600 = 300000 
Volume (ordinate cc,) = 2750 

Distance to centroid = 300000/2750 

= 109-1 ft 
Chainage of centroid = 350 - 109*1 
= 240-9 ft 

For area dbd, 

Area scaled = 1-9688 in 2 

.-. Haul = 1-9688 x 320000 = 630016 

Volume = (ordinate dd,) = 2750 

Distance to centroid = 229-1 ft 

Chainage of centroid = 650 + 229-1 

= 879-1 ft 

Average haul distance = 879-1 - 240-9 
= 638-2 ft 

Overhaul distance = 638-2 - 300 
= 338-2 ft 

.-. Overhaul = 338*2 x 2750 

= 344-5 station yards. 

N.B. Instead of the above calculation the overhaul can be obtained 
direct as the sum of the two mass-haul curve areas acc y and dbd, . 

. 300000 4 ' . 

Area acq = . station yd 

a aua 630016 „ 4 . 
Area aba^ = — jjnn station yd 

Total area = overhaul = . ftft = 344*5 station yards 




Proof 



Fig. 10.46 



Take any area cut off by a balancing line, Fig. 10.46. 

Let a small increment of area 8 A = (say) 1 yd 3 and length of 



552 


SURVEYING PROBLEMS AND SOLUTIONS 


haul be /. 






8A = 1 yd 3 x Z/100 station yd 




A = n x 1 yd x — 
n 




= Total volume x average haul distance, 




*. Area = Total Haul 



Exercises 10(c) (Earthwork volumes) 

15. Calculate the cubic contents, using the prismoidal formula of the 
length of embankment of which the cross-sectional areas at 50 ft in- 
tervals are as follows: 

Distance (ft) 50 100 150 200 250 300 
Area (ft 2 ) 110 425 640 726 1590 1790 2600 

Make a similar calculation using the trapezoidal method and ex- 
plain why the results differ. 

(I.C.E. Ans. 11688 yd 3 ; 12085 yd 3 ) 

16. The following notes were taken from the page of a level book: 

Reduced level Remarks 

45-85 At peg 10 

44-10 30 ft to right at peg 10 

44-75 30 ft to left at peg 10 

46-35 At peg 11 

42-85 30 ft to right at peg 11 

48-35 30 ft to left at peg 11 

46-85 At peg 12 

Draw cross-sections to a scale of 1 in. = 10 ft at pegs 10 and 
11, which are 100 ft apart on the centre line of a proposed branch rail- 
way, and thereafter calculate the volume of material excavated between 
the two pegs in forming the railway cutting. The width at formation 
level is 15 ft, and the sides of the cutting slope at V/2 horizontal to 
1 vertical. The formation level of each peg is 30*5 ft. 

(M.Q.B./M Ans. 2116 yd 3 ) 

17. A level cutting is made on ground having a uniform cross-slope of 
1 in 8. The formation width is 32 ft and the sides slope at 1 vertical 
to 1% horizontal. At 3 sections, spaced 66 ft apart, the depths to the 
centre line are 34, 28 and 20 ft. 

Calculate (a) the side widths of each section (b) the volume of 
the cutting. 

(N.R.C.T. Ans. 62-0; 96*6 ft; 53-3; 83-2 ft; 41*8, 65'2 ft; 

11600 yd 3 ) 



VOLUMES 553 

18. Calculate the volume in cubic feet contained between three suc- 
cessive sections of a railway cutting, 50 ft apart. The width of forma- 
tion is 10 ft, the sides slope 1 vertical to 2 horizontal and the heights 
at the top of the slopes in feet above formation level are as follows: 





Left 


Centre 


Right 


1st Cross-section 


13-6 


12-0 


14-0 


2nd Cross-section 


16-0 


15-5 


17-8 


3rd Cross-section 


18-3 


16-0 


16-0 






(N.R.C.T. 


Ans. 63 670 ft 3 ) 



19. The formation of a straight road was to be 40 ft wide with side 
slopes 1 vertically to 2^ horizontally in cutting. At a certain cross- 
section, the depth of excavation on the centre line was 10 ft and the 
cross-fall of the natural ground at right angles to the centre line was 

1 vertically to 8 horizontally. At the next cross-section, 100 ft away, 
the depth on the centre line was 20 ft and the cross-fall similarly 
1 in 10. 

Assuming that the top edge of each slope was a straight line, find 
the volume of excavation between the two sections by the prismoidal 
formula and find the percentage error that would be made by using the 
trapezoidal formula. 

(L.U. Ans. 11853 ft 3 12*5 %) 

20. A straight embankment is made on ground having a uniform cross- 
slope of 1 in 8. The formation width of the embankment is 30 ft and 
the side slopes are 1 vertical to V/z horizontal. At three sections 
spaced 50 ft apart the heights of the bank at the centre of the formation 
level are 10, 15 and 18 ft. Calculate the volume of the embankment 
and tabulate data required in the field for setting out purposes. 

(L.U. Ans. 2980 yd 3 ) 

21. Cross-sections at 100 ft intervals along the centre line of a pro- 
posed straight cutting are levelled at 20 ft intervals from -60 ft to 
+ 60 ft and the following information obtained: 

Distances (ft) -60 -40 -20 +20 +40 +60 

4-0 1-0 0-0 0-0 0-0 1-0 2-8 

100 12-9 8-6 5-0 3-0 2-0 3*0 6-0 

200 17-5 14-1 10-9 8-0 6-0 6-0 9-6 

300 21-8 17-7 14-4 11-3 9'7 9*7 11-0 

400 25-0 21-2 18-0 15*2 12*8 12-0 13*2 

(Tabulated figures are levels in feet relative to local datum). The for- 
mation level is zero feet, its breadth 20 ft, and the side slopes 1 ver- 
tical to 2 horizontal. Find the volume of excavation in cubic yards 
over the section given. (L D Ans 510 o y d 3 ) 



554 SURVEYING PROBLEMS AND SOLUTIONS 

22. A minor road with a formation width of 15 ft is to be made up a 
plane slope of 1 in 10 so that it rises at 1 in 40. There is to be no 
cut or fill on the centre line, and the side slopes are to be 1 vertical 
to 2 horizontal. Calculate the volume of excavation per 100 ft of road. 
Derive formulae for calculating the side-widths and heights and the 
cross-sectional area of a 'two-level' section. 

(N.R.C.T. Ans. 25 yd 3 /100 ft) 

23. The uniform slope of a hillside (which may be treated as a plane 
surface) was 1 vertically to 4 horizontally. On this surface a straight 
centre line AB was laid out with a uniform slope of 1 vertically to 

9 horizontally. With AB as the centre line a path with a formation 
width of 10 ft was constructed with side slopes of 1 vertically to 
2 horizontally. If the path was 500 ft in length and there was no cut 
or fill on the centre line, calculate the quantity of cutting in cubic feet. 

(I.C.E Ans. 2530 ft 3 ) 

24. The central heights of the ground above formation at three sections 
100 ft apart are 10, 12 and 15 ft and the cross-falls at these sections 
1 in 30, 1 in 40 and 1 in 20 (vertically to horizontally). If the forma- 
tion width is 40 ft and the side slopes 1 vertically in 2 horizontally, 
calculate the volume of excavation in the 200 ft length 

(a) if the centre line is straight, 

(b) if the centre line is an arc of 400 ft radius. 

(L.U. Ans. 158 270 ft 3 ; 158 270 ±1068 ft 3 ) 

25. The centre line of a highway cutting is on a curve of 400 ft radius, 
the original surface of the ground being approximately level. The cut- 
ting is to be widened by increasing the formation width from 20 to 30 ft, 
the excavation to be entirely on the inside of the curve and to retain 
the existing side slopes of 1V 2 horizontal to 1 vertical. If the depth of 
formation increases uniformly from 8 ft at ch. 600 to 17 ft at ch. 900, 
calculate the volume of earth to be removed in this 300 ft length. 

(L.U. Ans. 1302 yd 3 ) 

26. The contoured plan of a lake is planimetered and the following 
values obtained for the areas enclosed by the given underwater con- 
tours: 

Contour (ft O.D.) 305 300 295 290 285 

Area (ft 2 ) 38500 34700 26 200 7800 4900 

The surface area of the water in the lake is 40 200 ft 2 . The top 
water level and the lowest point in the lake are at 308-6 and 280*3 ft 
O.D. respectively. Find the quantity of water in the lake in millions of 
gallons. 

(L.U. Ans. 3-73 m. gal) 



VOLUMES 555 

27. The areas of ground within contour lines at the site of a reservoir 
are as follows: 

Contour in ft above datum Area (ft 2 ) 

400 505602 

395 442 104 

390 301 635 

385 232203 

380 94056 

375 56821 

370 34 107 

365 15834 

360 472 

Taking 360 ft O.D. as the level of the bottom of the reservoir and 
400 ft O.D. as the water level, estimate the quantity of water in gallons 
contained in the reservoir (assume 6*24 gal per ft 3 ). 

(Ans. 45 276500 gal) 

28. Describe three methods of carrying out the field work for obtaining 
the volumes of earthworks. 

Explain the conditions under which the 'end area' and 'prismoidal 
rule' methods of calculating volumes are accurate, and explain also the 
use of the 'prismoidal correction'. 

The areas within the contour lines at the site of a reservoir are as 
follows: 



Contour (ft) 


Area (ft 2 ) 


400 


5 120 000 


395 


4642000 


390 


4060000 


385 


3184000 


380 


2356000 


375 


1 765 000 


370 


900000 


365 


106000 


360 


11000 



The level of the bottom of the reservoir is 360 ft. Calculate (a) the 
volume of water in the reservoir when the water level is 400 ft using 
the end area method, (b) the volume of water in the reservoir using the 
prismoidal formula (every second area may be taken as a mid- area), and 
(c) the water level when the reservoir contains 300000000 gallons. 

(L.U. Ans. (a) 97-8925 m. ft 3 (b) 97-585 m. ft 3 388 ft) 
29. A square level area ABCD (in clockwise order) of 100 ft side is 
to be formed in a hillside which is considered to have a plane surface 
with a maximum gradient of 3 (horizontally) to 1 (vertically). 



556 SURVEYING PROBLEMS AND SOLUTIONS 

£ is a point which bisects the side AD, and the area ABE is to 
be formed by excavation into the hillside, whilst the area BCDE is to 
be formed on fill. The side slopes in both excavation and fill are to be 
1 to 1, and adjacent side slopes meet in a straight line. 

By means of contours at 2 ft intervals, plot the plan of the earth- 
works on graph paper to a scale of 50 ft to 1 inch. Hence compute the 
volume of excavation. 

(I.C.E. Ans. V ~ 880 yd 3 ) 

30. A road having a formation width of 40 ft with side slopes of 1 in 1 

is to be constructed. Details of two cross- sections of a cutting are as 

follows: 

Depth of Cutting Side Slope Limits (ft) 
Chainage (it) on Centre Line (ft) Left Right 

500 10-2 25-2 33-7 

600 6-0 22-0 28-5 

Assuming that these cross-sections are bounded by straight lines 
and that the undisturbed ground varies uniformly between them, com- 
pute the volume of excavation allowing for prismoidal excess. 

If instead of being straight, the plan of the centre line had been a 
circular curve of radius R with the centre of curvature on the right, 
how would this have been taken into account in the foregoing calcula- 
tions? Quote any formula that would have been used. 

(I.C.E. Ans. 1370 yd 3 ) 

31. A section of a proposed road is to run through a cutting from chain- 
age 500 to 900, the formation level falling at 1 in 200 from chainage 
500. The formation width is to be 30 ft and the side slopes are to be 

1 vertical to 2 horizontal. The original ground surface is inclined 
uniformly at right-angles to the centre line at an inclination of 1 in 10. 
With the information given below, calculate the volume of excava- 
tion in cubic yards, using the prismoidal formula. 

->, . ^ .. T , Ground Level at 

Chainage Formation Level « ^ T • 

Centre Line 

500 44-25 ft 51-11 

600 50-82 

700 50-93 

800 51-09 

900 50-77 

(I.C.E. Ans. 5474 yd 3 ) 

32. A road of 40 ft formation width is to be constructed with side 
slopes of 1 (vertical) to V-/i (horizontal) in excavation and 1 (vertical) 
to 2 (horizontal) in fill. Further details of two cross-sections are given 



VOLUMES 557 

below where the cross fall of the undisturbed ground is 1 (vertical) to 
r (horizontal). 



Chainage (ft) 


Ground level on 


Formation Level 


Centre-line (ft above datum) 


(ft above datum) 


400 


171-6 


166-6 4 


500 


170-2 


168-0 6 



Assuming the road is straight between these two sections, compute 
the volumes of excavation and fill in 100 ft length neglecting prismoi- 
dal excess. 

(I.C.E. Ans. 819 yd 3 cut, 11 yd 3 fill) 

33. On a 1000 ft length of new road the earthwork volumes between 
sections at 100 ft intervals are as follows, the excavation being taken 
as positive and filling as negative: 

Section No 01234 56 7 8 9 10 

Vol. (1000 yd 3 ) 3-7 9-1 15-0 13-9 6-4 1-4 -5-6 -19-4 -18-9-5-6 
Draw the mass- haul curve and find 
(i) the volume to be moved under the terms of the free-haul limit 

of 300 ft, 
(ii) the volume to be moved in addition to (i), 
(iii) the number of station-yards under (ii) where 1 station-yard 

equals 1 cubic yard moved 100 ft, 
(iv) the average length of haul under (ii). 

(L.U. Ans. 7500 yd 3 ; 42 000 yd 3 ; 23500 station yds; 560 ft) 

34. The following figures show the excavation (+) and filling (-) in 
cubic yards between successive stations 100 ft apart in a proposed 
road. 





1 


2 


3 


4 5 


6 


+ 1500 


+ 1100 


+ 500 


+ 100 - 100 


-1000 


7 


8 


9 


10 11 


12 


-2200 


-2500 


-1600 


-400 +1800 


+ 2800 



State which of the following tenders is the lower and calculate the 
total mass haul in the 1200 ft length: 

(a) Excavate, cart and fill at 9/6 per yd 3 . 

(b) Excavate, cart and fill at 9/- per yd 3 , with a free-haul limit of 
400 ft, plus 1/- per station yard for hauling in excess of 400 ft. 

(1 station yard = 100 ft x 1 yd 3 ). 

(L.U. Ans. 28 500 station yards; (a) £3700, (b) £3750) 

35. Volumes in yd 3 of excavation (positive) and fill (negative) between 
successive sections 100 ft apart on a 1300 ft length of a proposed 
railway are given in the following table: 



558 SURVEYING PROBLEMS AND SOLUTIONS 

Section 1 2 3 4 5 6 

Volume -1000 -2200 -1600 -500 +200 +1300 

7 8 9 10 11 12 13 

+ 2100 +1800 +1100 +300-400 -1200 -1900 

Draw a mass haul curve for this length. If earth may be borrowed 
at either end, which alternative would give the least haul ? Show on 
the diagram the forward and backward free-hauls if the free-haul limit 
is 500 ft, and give these volumes. 

(L.U. Ans. Borrow at end, 1150 ft; 2900 yd 3 ; 2400 yd 3 ) 

36. The volumes in yd 3 between successive sections 100 ft apart on 
a 900 ft length of a proposed road are given below; excavation is 
shown positive and fill negative 

Section 012 3 4 56 78 9 

Volume +1700 -100 -3200 -3400 -1400 +100 +2600 +4600+1100 

Determine the maximum haul distance when earth may be wasted 
only at the 900 ft chainage end. 

Show and evaluate on your diagram the overhaul if the free-haul 
limit is 300 ft. 

(L.U. Ans. 510 ft; 4950 station yards) 

Bibliography 

bannister, a. and RAYMOND, s., Surveying (Pitman) 

DAVIES, R.E. FOOTE, F.S., and KELLY, J.W., Surveying Theory and 

Practice (McGraw Hill) 
CLARK, D., Plane and Geodetic Surveying, Vol.1 (Constable) 
THOMAS, W.N., Surveying (Edward Arnold) 
CURTIN, w. and LANE, R.F., Concise Practical Surveying (English 

Universities Press) 
RITTER-PAQUETTE, Highway Engineering, 2nd Ed. (Ronald Press) 
SHARMA, R.c. and SHARMA, S.K. Principles and Practice of Highway 

Engineering (Asia Publishing House) 
MINISTRY OF TECHNOLOGY, Changing to the Metric System (H.M.S.O.) 



11 



CIRCULAR CURVES 



11.1 Definition 

The curve can be defined by (a) the radius R or (b) the degree of 
the curve D. D can be expressed as the angle at the centre of the 
curve subtended by (i) a chord of 100 ft or (ii) an arc of 100 ft (the 
former is more generally adopted). 
In Fig. 11.1, 



(11.1) 



. , _ i x 100 50 
Sm ^ " 2 R " R 




r R- 50 




sin^D 




D is small, 




sin | D = % D radians 




206265x50 
K = ~ 


5730 


\D° x3600 


D° 



(11.2) 




100' 



Fig. 11.1 



11.2 Through Chainage 

Through chainage represents the length of road or rail from some 
terminus (it does not necessarily imply Gunter's chain — it may be the 
engineer's chain). Pegs are placed at 'stations' (frequently at 100 ft 
intervals) and a point on the construction can be defined by reference 
to the * station'. 

When a curve is introduced, Fig. 11.2, the tangent point T, is said 
to be of chainage 46 + 25, i.e. 4625 ft from the origin. If the length of 
the curve was 400 ft, the chainage of T z would be expressed as 
(46 + 25) + (4 + 00), i.e. 50 + 25. 



Origin of system 
12 3 



T, 46+25 
42 43 44 45 46/47 



Chainage of 7i — 46+25 
Length of arc » 400ft 
Chainage of T, ■ 50+25 




Fig. 11.2 Through chainage 

559 



560 SURVEYING PROBLEMS AND SOLUTIONS 

11.3 Length of Curve (L) 

L = 27TR * §fo (11 - 3) 

or L = R.6 Tad (11.4) 

The second formula is better. 



Example 11.1 If 6 = 30° 26 and R = 100 ft, 

26 
2 x 3-142 x 100 x 30^ 
60 
Length of arc L = — — = 53 '12 ft 

or L = 100 x 30°26; ad 

- 100 x 0*53116 = 53-12 ft 



11.4 Geometry of the Curve 

In Fig. 11.3, T, and T 2 are tangent 
points, / is the intersection point, 
is the deflection angle at inter- 
section point, /T, = IT 2 - tangent 

length = Rtan±<j>. 

T, T 2 = Long chord = 2 . TX 

= 2Ksin^c£ (11.5) 

T,A = chord c = 2R sin a (11.6) 

(if c < R/20 and a is small, 
arc ~ chord, i.e. sin a = a rad ) 




Fig. 11.3 Geometry of the curve 



c = 2Ra Tad (11.7) 

Deflection angle a rad = S- (11.8) 

206 265 c 

a sec= — 2£ (119) 

206265c 

a — = 2^60" (U - 10) 

. l-Z^£ (ii.iD 



CIRCULAR CURVES 



561 



10 



R sec \ <f> 



(11.12) 



IP = 10 - PO = R sec±<f> - R = R(sec\cf> - 1) (11.13) 
PX = PO - XO = R - flcosl0 = /?(1- cos^) (11.14) 

= R versine^0. (11.15) 

11.5 Special Problems 
11.51 To pass a curve tangential to three given straights (Fig. 11.4) 



(1) + (2) 



i.e 




R = T y Xcot±j> 

= (^ + z)cot|0 = (* 2 + y)cot^<£ 

= {{(x + z-y) + y)cot^<£ 

= ^(x + y + z) cot^<£ 

R = 5cot^0 where s = \ perimeter of A XYZ 



(11.16) 



562 SURVEYING PROBLEMS AND SOLUTIONS 

Alternative solutions: 

(a) Area XOY = \Rz 

XOZ = ±Ry 

YOZ = ±Rx 

XYZ = areas (XOY + XOZ - YOZ) 

= \R(y+z-x) 

= ±R(x + y+z)- \R(x+ x) 

= Rs - Rx 

= R(s - x) 

area A XYZ 
:. R = . (11.17) 

S - X 

(b) If angles a and /3 are known or computed, 

YP = R tan a/2 
PZ = R tan 0/2 
.-. YZ = YP + PZ = K(tan a/2 + tan /S/2) 

YZ 

R = tana/2 + tan /S/2 (1118) 

Example 11.2 The co-ordinates of three stations A, B and C are as 
follows :- 

A E 1263-13 m N1573'12m 
B E 923*47 m N 587'45m 
C E 1639-28 m N 722-87 m. 

The lines AB and AC are to be produced and a curve set out so 
that the curve will be tangential to AB, BC and AC. Calculate the 
radius of the curve. 

In Fig. 11.5, 

,923-47- 1263-13 



Bearing AB = tan" 



587-45 - 1573-12 



, -, -339-66 

= tan ' 

- 985-67 
= S19°00' 50" W 
= 199° 00' 50". 




Fig. 11.5 



CIRCULAR CURVES 563 

Length AB = 985*67 sec 19° 00' 50" = 1042-55 

n- Q •„ An . _,1639-28- 1263*13 _, 376*15 

Bearing AC = tan ' = tan ' 

722-87 - 1573-12 - 850-25 

= S23°51'52"E = 156°08'08". 



Length AC = 850*25 sec 23° 51' 52" = 929*74 

Bearing EC - tan -, 1639*28 - 923*47 = ^-,715*81 
722-87 - 587-45 135-42 

= N79°17'14"E = 079° 17' 14" 

Length BC = 135*42 sec 79° 17' 14" = 728*51 

= 180 - (19° 00' 50" + 23° 51' 52") = 137°07'18" 

To find the radius 
Using Eq. (11.16), 

R = {(AB+BC+ AC) cot \4> 

= ^(1042*55 + 728*51 + 929*74) cot 68° 33' 39" 

= 1350*40 cot 68° 33' 39" = 530*2 m. 

Alternatively, by Eq. (11.17), 

R = V / fc( 5 - a >( s - b H s - c >i 
s — a 

= Vl 1350*40(1350*40 - 728*51)(1350*40 - 929*74) (1350*40 - 1042*54)} 

1350*40 - 728*51 
= 530*2 m. 

Alternatively, by Eq. (11.18), 

a = 079° 17' 14" - 019° 00' 50" = 60° 16' 24" 
)8 = 336° 08' 08" - 259° 17' 14" = 76° 50' 54" 

check <f> = 137° 07' 18" 

o = 728*51 

tan 30° 08' 12"+ tan38°25'27" " ^L£B^ 

11.52 To pass a curve through three points (Fig. 11.6) 

In Fig. 11.6, 

angle COA = 20 
angle XOC = 180 - 

XC = ±AC = R sin (180-0) 



564 SURVEYING PROBLEMS AND SOLUTIONS 

AC 



sin i 



= 2R, 









\AC 






i.e. 


R = 




(11.19) 




sin 6 


Similarly, the full sine 


rule is 


written 










AB 
sinC 


BC 

sin A 


AC 
sin B 


= 2R 












(11.20) 




Example 11.3 The co-ordinates of two points B and C with respect 
to A are 

B 536*23 mN 449*95 m E 

C 692-34 mN 1336'28mE. 

Calculate the radius of the circular curve passing through the 

three points. 

449*95 
Bearing AB = tan" 1 ——— = N40°E 
boo'Zo 

Bearing BC = tan^ fff'ff = N80°E 
156*11 

.-. angle ABC = 6 = 180 + 40 - 80 = 140°. 

Length AC = V(692*34 2 + 1336-28 2 ) = 1505 



m 



1505 752*5 _ 

2 sin 140° " sin 40 " t±I}LL 



Example 11 .4 In order to find the radius of an existing road curve, 
three suitable points A, B, and C were selected on the centre line. 
The instrument was set up at B and the following tacheometrical read- 
ings taken on A and C, the telescope being horizontal and the staff 
held vertical in each case. 

Staff at Horizontal angle Collimation Stadia 
A 0°00' 4*03 5*39/2*67 

C 195° 34' 6*42 8*04/4*80. 



CIRCULAR CURVES 



565 



If the instrument had a constant multiplier of 100 and an additive 
constant of zero, calculate the radius of a circular arc ABC. 

If the trunnion axis was 5*12 ft above the road at B, find the 
gradients of AB and BC. (L.U.) 

In Fig. 11.7, 

AB = 100(5-39-2-67) = 272 ft 

BC = 100(8-04-4-80) = 324 ft. 

In triangle ABC, 

. A-B 324- 272 „ 195°34'-180 

tan — - — = tan 

2 324 + 272 2 

52 tan7°47' 



596 




Fig. 11.7 



A-B 



R = 



2 


V tJ. 


A + B 
2 


7° 47' 


A = 


8° 28' 


B = 


7° 06' 


BC 

sin A 


2R 


324 


1 1 fifi»3 ff 


2 sin 8° 


28' " 11003ft - 



Difference in height A-B, 5*12 - 4*03 = 1*09 ft. 
Gradient AB, 1*09 in 272 = 1 in 240. 
Difference in height B - C, 6'42 - 5*12 = 1*30 ft. 
Gradient BC, 1 -30 in 324 = 1 in 249. 
If the gradients are along the arc, 



= 2 sin -1 



324 



= 16° 56' 



2 x 1100-3 

Length of arc AB = 1100'3 x 16° 56 ' rad - 272 '7 ft 

6> 2 = (2x15° 34')- 16° 56' = 14° 12' 
Length of arc BC = 1100-3 x 14° 12; ad = 352'2 ft. 

Gradients along the arc are 1 in 251; 1 in 250. 



566 SURVEYING PROBLEMS AND SOLUTIONS 

Exercises 11 (a) 

1. It is required to range a simple curve which will be tangential to 
three straight lines YX, PQ, and XZ, where PQ is a straight, joining 
the two intersecting lines YX and XZ. Angles YPQ = 134° 50' ; 
YXZ = 72° 30' ; PQZ = 117° 40" and the distance XP = 5'75 chains. 

Compute the tangent distance from X along the straight YX and 
the radius of curvature. 

(I.C.E.. Ans. 8*273 chains; 6*066 chains) 

2. A circular road has to be laid out so that it shall be tangent to 
each of the lines DA, AB, and BC. 

Given the co-ordinates and bearings as follows, calculate the 
radius of the circle. 

Latitude Departure 

A - 29*34 m - 128*76 m 
B - 177-97 m - 58-39 m 

Bearing DA 114° 58' 10" 
CB 054° 24' 10" 

(Ans. 137-48 m) 

3. A circular curve is tangential to three straight lines AB, BC, and 
CD, the whole circle bearings of which are 38°, 72° and 114° respec- 
tively. The length of BC is 630 ft. Find the radius and length of the 
curve and the distances required to locate the tangent points. 

Also tabulate the data necessary for setting out the curve from the 
tangent point on BC with 100 ft chords and a theodolite reading to 30" 
with clockwise graduations. 

(L.U. Ans. K = 913-6ft; 1211'9ft; 279-3 and 350'7ft) 

4. The co-ordinates of two stations Y and Z in relation to a station 
X are as follows: 





E(m) 


N(m) 


Y 


215 


576 


Z 


800 


-750 



Find by construction the radius of a circular curve passing through 
each of the stations X, Y, and Z . 

(Ans. 790 m) 

5. Three points A, B, and C lie on the centre line of an existing 
mine roadway . A theodolite is set up at B and the following observa- 
tions were taken on to a vertical staff. 

Staff at Horizontal circle Vertical circle c . ,. ~ ... 

Stadia Collimation 

A 002° 10' 20" +2° 10' 6*83/4-43 5*63 

C 135° 24' 40" -1°24' 7-46/4-12 5'79 



CIRCULAR CURVES 



567 



If the multiplying constant is 100 and the additive constant zero, 
calculate: 

(a) the radius of the circular curve which will pass through A, B 
and C, 

(b) the gradient of the track laid from A to C if the instrument 
height is 5-16 ft. 

(R.I.C.S./M Ans. 362-2 ft; 1 in 33-93) 



11.53 To pass a curve through a given point P 

(a) Given the co-ordinates of P relative to I, i.e. I A and AP 
(Fig. 11.8) 



I. 

♦/ 2) VL N 


0* 

Jfl) \r 




JT^^^Sv 


\ 

X 

X 

\ 
\ 

N: 


/ 

f 
* 

/ 

/ 
/ 

/ 

t 



Fig. 11.8 
In Fig. 11.8, 

Assume BC passing through P is parallel to 7, T z . 
AB = AP cot \<f> 
BP = AP cosec \<f> = QC 

BX = (AB + AI) cos \<f> 

PX = BX - BP 

BQ = BX + PX 

T,B = y/(BP x BQ) (intersecting tangent and 

T,/ = BI ±T,B (i.e. T, 1 ^ = T, B) 
R = T x l cot - <f> or T, 1 / cot \<f> 



secant) 



= (AB + AI ±T,B) cot \<f> 



(11.21) 



568 SURVEYING PROBLEMS AND SOLUTIONS 

(b) Given polar co-ordinates of point P from intersection I (Fig. 11. 9) 

In Fig. 11. 9, 

a = 90-(f +0) 

01 = R sec|<£ 

01 _ Rsec\cf> 
OP R ' 

also by sine rule 

sin (a + jS) 
sin a 
••• sin (a + /3) = sin a sec ^<f> 

[ = sinl80-(a + i 8)] (11.22) 

Thus there are 2 values for /3. 




R 



IP 



sin a sin /3 

R = IP sin a cosec j8 



Fig. 11.9 



(2 values) 



(11.23) 




Fig. 11.10 



CIRCULAR CURVES 



569 



If a = 0, in Fig. 11.9, i.e. P lies on line 10, 

1 



01 = R sec 10 = IP ± R 



R(sec\c/> +1)= IP 



R = 



IP 



secl<£ ± 1 



If 0=90°, in Fig. 11.10, 
T,0 = R 
QP =!,/ = /? sinS - R tan^c£ 

sin 8 = tan^<£ 

7,(3 = IP = R - RcosS 
= K(l -cosS) 

verso 
As 5 has 2 values there are two values of R . 



(11.24) 



(11.25) 



Example 11.5 A circular railway curve has to be set out to pass 
through a point which is 40 m from the intersection of the straights and 
equidistant from the tangent points. 

The straights are deflected 
through 46° 40 ' . Calculate the radius 
of the curve and the tangent length. 

Taking the maximum radius only, 
in Fig. 11.11, 

01 = OP + PI 

= R + x = R sec^<£ 

.*. R(l - sec|0)= - x 
i.e., Eq. (11.24), 

x 



R 



40 



secl<£ - l 



1-089 07- 1 




449-1 m. 



Fig. 11.11 



TJ = T 2 I = R tan-^0 

= 449-1 tan 23° 20' 



193-7 m. 



570 



SURVEYING PROBLEMS AND SOLUTIONS 



Example 11.6 The bearings of two lines AB and BC are 036° 36' 
and 080° 00' respectively. 

At a distance of 276 metres from B towards A and 88 metres at 
right-angles to the line AB, a station P has been located. 

Find the radius of the curve to pass through the point P and also 
touch the two lines. (R.I.C.S./M) 




Taking the maximum 
radius only 



In Fig. 11.12, 



Fig. 11.12 



cf> = 080° 00' - 036° 36' = 43° 24' 
In triangle GLP, 

LG = LP cot \<$> = 88cot21°42' = 221*13 m 

GP = LPcosec±(f> = 88cosec21°42' = 238*00 m 

GX = (LG + LB) cos^<£ = (221*13 + 276*00) cos 21° 42' = 461*90m 

PX = GX - GP = 461*90 - 238*00 = 223*90 m 
GQ = GX + PX = 461*90 + 223*90 = 685*80 m 

T,G= y/(GP x GQ) = V(238*00 x 685*80) = 404*01 m 
T,B = GB + T,G = BL + LG + GT, 

= 276*0 + 221*13 + 404*01 = 901*14 m 
R = T,B cot|0 = 901*14 cot 21° 42' 
= 2264*5 m. 



CIRCULAR CURVES 571 

Alternative solution 

6 = tan"' 88/276 = 17° 41' 

a = 90 - (±<f> + 6) = 90 - (21° 42' + 17° 41') 

= 50° 37' 
x = 276 sec0 = 289*69 m 
Then, from Eq. (11.22), 

sin(a+/8) = sinasec^0 

= sin 50° 37 'sec 21° 42' 
a+/3 = 56° 17' 30" 
a = 50° 37' 00" 
.-. B = 5° 40 '30" 
Therefore radius R = x sin a cosec j8 

= 289-69 sin 50° 37' cosec 5° 40' 30" 
= 2264*5 m 



Exercises 11. (b) (Curves passing through a given point) 
(N.B. In each case only the maximum radius is used) 

6. In setting out a circular railway curve it is found that the curve 
must pass through a point 50 ft from the intersection point and equi- 
distant from the tangents. The chainage of the intersection point is 
280 + 80 and the intersection angle (i.e. deflection angle) is 28°. 

Calculate the radius of the curve, the chainage at the beginning 
and end of the curve, and the degree of curvature . 

(I.C.E. Ans. 1633 ft; 276 + 73; 284 + 73; 3° 30') 

7. Two straights intersecting at a point B have the following bear- 
ings: BA 270°; BC 110°. They are to be joined by a circular curve, 
but the curve must pass through a point D which is 150 ft from B and 
the bearing of BD is 260°. Find the required radius, the tangent 
distances, the length of the curve and the deflection angle for a 100 ft 
chord. 

(L.U. Ans. 3127*4 ft; 551*4 ft; 1091*7 ft; 0°55') 

8. The co-ordinates of the intersection point / of two railway 
straights, AI and IB, are 0,0. The bearing of AI is 90° and that of 
IB is 57° 14'. If a circular curve is to connect these straights and if 
this curve must pass through the point whose co-ordinates are 

- 303*1 ft E and + 20*4 ft N, find the radius of the curve. 



572 SURVEYING PROBLEMS AND SOLUTIONS 

Calculate also the co-ordinates of the tangent point on Al and the 
deflection angles necessary for setting out 100 ft chords from this 
tangent point. What would be the deflection angle to the other tangent 
point and what would be the final chord length? 

(L.U. Ans. 2000ft; 588*0 ft E Oft N) 

9. A straight BC deflects 24° right from a straight AB. These are 

to be joined by a circular curve which passes through a point D 200 ft from 
from B and 50 ft from AB. 

Calculate the tangent length, the length of the curve, and the def- 
lection angle for a 100 ft chord. 

(L.U. Ans. 818-6; 806*5; 0° 45') 

10. A right-hand circular curve is to connect two straights Al and IB, 
the bearings of which are 70° 42' and 130° 54' respectively. The curve 
is to pass through a point X such that IX is 132*4 ft and the angle 
AIX is 34° 36'. Determine the radius of the curve. 

If the chainage of the intersection point is 5261 ft, determine the 
tangential angles required to set out the first two pegs on the curve at 
through chainages of 50 ft. 

(L.U. Ans. 754*1 ft; 0° 59' 40"; 2° 53' 40") 

11. The following readings were taken by a theodolite stationed at the 
point of intersection / of a circular curve of which A and B are re- 
spectively the first and second tangent points 

A 14° 52'; B 224° 52'; C 344° 52' 
It is required that the curve shall pass through the point C, which 
is near the middle of the curve, at a distance of 60 ft from /. 

(a) Determine the radius of the curve. 

(b) Calculate the running distances of the tangent points A and 
B and the point C, the distance at / being 200 + 72, in 
100 ft units. 

(c) Show in tabular form the running distances and tangential 
angles for setting out the curve between A and C. 

(L.U. Ans. 1181ft; 197 + 56; 203 + 74; 200 + 22) 



11.54 Given a curve joining two tangents, to find the change required 
in the radius for an assumed change in the tangent length 
(Fig. 11.13) 



In Fig. 11.13, 










R 2 -R, 


= (*a-',)cotJL0 


(11.26) 




o,o 2 


= (R 2 - /?,) secj0 


(11.27) 



CIRCULAR CURVES 



573 



A, A, — 



R, = 



IX 2 - IX, 

(/0 2 - R 2 ) - (70, - /?,) 

(R 2 sec {$ - R 2 ) - (K, sec \<f> - /?,) 

/?2(sec|0 - 1) - R,(sec\cf> - 1) 

(/? 2 -/?,)(secl0-l) 



X,X 2 



sec-i0- 1 



+ «i 



(11.28) 
(11.29) 




Fig. 11.13 

Example 11.7 (a) A circular curve of 2000 ft radius joins two points 
A and C which lie on the two straights AB and BC. If the running 
chainage values of A and C are 1091ft and 2895 ft respectively, 
calculate the distance of the midpoint of the curve from B. 

(b) If the minimum clearance value of the curve from B is to be 
200ft, what radius would be required for the curve and what would be 
the chainage value for the new tangent points? 

(R.I.C.S.) 

(a) In Fig. 11.14, 

Length of curve - 2895 - 1091 
= 1804 ft 



= R<k 



rad 



0. 



rad — 



arc 
~R 



574 



SURVEYING PROBLEMS AND SOLUTIONS 



- 1804 

" 2ooo 

<f> = 51° 41' 
In triangle T^BO u 
BO, = R sec^0 

= 2000sec25°50'30" 
= 2222-22 ft 

BX, = 2222-22 - 2000 
= 222-22 ft 

In triangle T 3 BO z , 
By Eq. (11.29), 



8/51*41' 




1187-86 r , 



j 2811-46 



Fig. 11.14 



/?, = R s 



X,X 2 



sec -^ <f> - 1 



- 2000 -•• 



222-22 - 200 



1-1111-1 
- 2000 - 200 
= 1800 ft 



(b) By Eq. (11.27), 



0,0 2 = (R z - /?,) sec ±cf> 

= (2000 - 1800) sec 25° 50' 30" 
= 222-22 ft 
TJ 3 = 2 P = 0,0 2 sin\c/> 

= 222*22 sin 25° 50 '30" 
= 96*86 ft 



Alternatively, by Eq. (11.26), 



e 2 - ',) = ( R z " *,) tan I <f> 
= 200 tan 25° 50' 30" 
= 96-86 ft 

Chainage T 3 = T, + (t 2 _ ^) 
= 1091 + 96*86 
= 1187*86 ft 



CIRCULAR CURVES 



575 



Length of arc = 1800 x <f> 

= 1800 x 0*902 

= 1623-60 

.'. Chainage T A = 1187-86 + 1623*60 

= 281 1-46 ft 

11.6 Location of Tangents and Curve 

// no part of the curve or straights exists, setting out is related to 
a development plan controlled by traverse and/or topographical detail . 

// straights exist 

(1) Locate intersection point /. 

(2) Measure deflection angle <£. 

(3) Compute tangent length if radius R is known . 

(4) Set off tangent points T, and T z . 

// the intersection point is inaccessible 

Select stations A and B on the straights, Fig. 11.15. 



Measure a and /3; AB. 
Solve triangle AIB. 

AI = AB sin /8 cosec <f> 
BI = AB sin a cosec <f> 
TJ = T 2 I = R tan|0 

/. T,A = T,/ - AI 
T 2 B = T,/ - Bl 




Fig. 11.15 



For through chainage 

Chainage of A known 
Chainage of / = Chainage A + AI 
Chainage of T, = Chainage / - T,/ 
Chainage of T z = Chainage T, + arc T, T 2 



576 



SURVEYING PROBLEMS AND SOLUTIONS 



// the tangent point is inaccessible 

If T, is not accessible set out where possible from T 2 , Fig. 11.16. 
A check is possible by selecting station B on the curve and checking 
offset AB. 




Fig. 11.16 
N.B. To locate the curve elements a traverse may be required joining 
the straights. It is essential to leave permanent stations as reference 
points for ultimate setting out. 

11 .7 Setting out of Curves 
11 .71 By linear equipment only 

(a) Offsets pom the long chord (Fig. 11.17) 




If, in Fig. 11.17, the chord is sub divided into an even number of 

equal parts, the offsets \h z etc. can be set out. Each side of the mid- 
point will be symmetrical. 

Generally, (R - yf = R z - x z 

y = R - y/(R 2 - x z ) (11.30) 

y 2 = R - y/(R z -xl) (11.31) 

.-. K - y - y 2 (11.32) 



N.B. 



If <f> is known, y = R(l - cosi<£), i.e. y = R versine^. 



Example 11.8 A kerb is part of a 100ft radius curve. If the chord 
joining the tangent points is 60ft long, calculate the offsets from the 

chord at 10ft intervals, Fig. 11.18. 



CIRCULAR CURVES 



577 




Fig. 11.18 
As above (Eq. 11.30), 

h 3 = y = R -y/(R 2 -x 2 ) 

= 100 - V(100 2 -30 2 ) 

= 100 - V(100 - 30) (100 + 30) 

= 4-61 . 

By Eqs.11.31/11.32, h 2 = h 3 - y 2 

y z = 100 - V(100 2 - 10 2 ) 

= 0-50 

h 2 = 4*61 - 0*50 - 4J1 

ft, = h 3 - y y 

y, = 100 - V(100 2 -20 2 ) 

= 2-02 

ft, = 4-61 - 2-02 = 2-59 . 

(b) Offsets from the straight (Fig. 11.19) 





Fig. 11.19 
As above, 
i.e., by Eq. (11.30), 
Alternatively, 



Fig. 11.20 



(R - xf = R z - y 2 
x = R - y/(R 2 -y 2 ) 



sin a = 



If a is small, c ~2±y. 



2R 
2R 



2R 



(11.33) 
(11.34) 

(11.35) 



578 



SURVEYING PROBLEMS AND SOLUTIONS 



(c) Offsets from the bisection of the chord (Fig. 11.20). 
From Eq. (11.30), AA< _ R _^ [Rl _(c^ x 

Alternatively, 

AA, = R(l -cos-^) 

BB, = K(l -cos{<£) 

CC, = RQ. -cos^) 

(d) Offsets from the bisection of successive chords; centre of curve 
fixed (Fig. 11.21). 

As above, 



sin a 



c 
2R 




Fig. 11.21 



Assuming equal chords, 








Tt = Aa 


= Bb = Dd etc. 




= K(l 


- cos a) = R 


vers a 


Alternatively, 


Tt = R 


-**'-(§)" 


} 




Lay off T,x = 


c 
2 






Lay off xA = 


Tt 






Aa = 


Tt along line 


AO 




T, B on line Ta produced 




Bb = 


Tt along line 


BO 


(e) Offsets from 


chords produced. 




(i) Equal chords (Fig. 11.22) 








A Z A = 


c sin a 





= C X 



2R 2R 



(Eq. 11.34) 



CIRCULAR CURVES 



579 



If a is small, A 2 A ^ A X A. This 
can be controlled if the length of 
chord is limited, 



i.e. 



R 
C< 20 



B,B = 2A X A if a is small. 

•»2 



B,B = 



R 



(11.36) 




(ii) Unequal chords (through chainage) 




Fig. 11.23 



In Fig. 11.23, 



Offset from sub-chord X = A X A = —^ 

Offset from 1st full chord 2 = B,B = B X B 2 + B 2 B 



(Eq. 11.34) 



C 1 C 2 f|_ C 2 ( C 1 + C 2 ) 

2/2 + 2/2 " 2/2 

(11.37) 



580 



SURVEYING PROBLEMS AND SOLUTIONS 



Offset from 2nd full chord = 3 = D, D = D y D 2 + DJ) 

= 2D,D = 2B.B 



By Eq. (11.37), 



Q, = 



c 3 ( c 2 + c 3 ) 
2K 



(11.38) 



but c 3 = c 2 



Generally, 



a = 



2/? 

Cn(Cn + C n -,) 



2i? 



(11.39) 



11.72 By linear and angular equipment 

(a) Tangential deflection angles (Fig. 11.24) 



By Eq. (11.33), sin a 


c 
= 2R 


If 


a is small ( c < 


R/20) 




sin a 


~ a rad 


.-. 


206 265 c 

a sec ~ 2R 


(11.40) 




206 265 c 


1718-8 c 




2R x 60 


R 
(11.41) 




Fig. 11.24 



4> 
For equal chords 2 a = — (11.42) 

n 

where n = no. of chords required, 

For sub-chords, 

sub- chord 




a. 



ax 



standard chord 



(11.43) 






Fig. 11.25 



(b) Deflection angles from chord produced (Fig. 11.25) 

The theodolite is successively moved round the curve. This method 
is applicable where sights from T, are restricted. It is the method 
applied underground. 

11.73 By angular equipment only (Fig. 11.26) 

Deflection angles are set out from each tangent point, e.g., A is 
the intersection of a from T, with 3 a from T 2 . 



CIRCULAR CURVES 



N.B. na = ±<f> 

Generally, if the deflection angle from T. is a. 



581 
(11.44) 



the deflection angle jS from T 2 = 360 - l<f> + a, (11.45) 





Fig. 11.26 

Example 11.9 In a town planning scheme, a road 30 ft wide is to in- 
tersect another road 40 ft wide at 60°, both being straight. The kerbs 
forming the acute angle are to be joined by a circular curve of 100 ft 
radius and those forming the obtuse angle by one of 400 ft radius. 

Calculate the distances required for setting out the four tangent 
points. 

Describe how to set out the larger curve by the deflection angle 
method and tabulate the angles for 50 ft chords. (L.U.) 



b d 



*■ c 




\ 

1' 

Fig. 11.27 
7, X = T Z X = 400 tan 30 = 230-94 ft 
T 3 Z = I 4 Z = 100 tan 60 = 173-21 ft 



In triangle XYb, 
20 



XY 



= YZ = 23-09 



sin 60 
Xb = 20 tan 30 = Yc = 11-55 



582 SURVEYING PROBLEMS AND SOLUTIONS 

In triangle BYa, 

15 

BY = — = 17-33 

sin 60 

aY = Bd = 15 tan 30° = 8-66 

Distances to tangent points measured along centre lines: 

AB = T X X+XY - aY = 230-94 + 23-09 - 8-66 = 245-37 ft 

BC = T A Z + XY + aY » 173-21 + 23-09+ 8-66 = 204-96 ft 

BD = T Z X -Xb +BY = 230-94 - 11-55 + 17-33 = 236-72 ft 

BE = T 3 Z + Xb + BY = 173-21 + 11-55 + 17-33 = 202-09 ft 

Deflection angles, 50 ft chords 

50 

sina i = « 7^ 

1 2 x 400 
a, = 3°35' 



By approximation, 



206 265x50 __ 
a_ = -— — : = 12892 sec 



2x400 



= 3°34'52" 



a, = 3°35' 

a 2 = 7°10' 

a 3 = 10°45' 

a 4 = 14°20' 

a 8 = 17°55' 

a 6 = 21°30' 

a 7 = 25°05' 

a 8 = 28°40' 

a 9 = 30°00' (sub-chord = 2^sinl°20' 

= 2x400sinl°20' 
= 18-62ft) 



Example 11.10 A curve of radius 1000 ft is to be set out connecting 
2 straights as shown in Fig. 11.28. 

Point X is inaccessible and BC is set out and the data shown 
obtained. Assuming the chainage at B is 46 + 47*3 ft, calculate 
sufficient data to set out the curve by deflection angles from the tan- 
gent by chords of 50 ft based on through chainage. 



CIRCULAR CURVES 



583 




0902-69} 7i / 



950) 

T, (599709) 



Fig. 11.28 
(derived values in brackets) 



In triangle BXC, (Fig. 11.28) 



BX = 1050-4 sin 54°30' cosec60° = 987-44 ft 
XC = 1050-4 sin 65°30' cosec60° = 1103-69ft 



In triangle OT^X, 








Tangent length 


T,X 


= 


1000 tan 60° 






= 


1732-05 ft 


.*. 


BT, 


= 


T,X- BX 






= 


1732-05 - 987-44 






= 


744-61 ft 


Similarly 


CT 2 


= 


1732-05 - 1103-69 






= 


628-36 ft 


Chainage 


r, 


= 


4647-30 - 744-61 






= 


3902-69 ft 


Length of curve 




= 


R <£ = 1000 x 2-094 






= 


2094-40 ft 


Chainage 


T z 


= 


3902-69 + 2094-40 






= 


5997-09 ft 



584 SURVEYING PROBLEMS AND SOLUTIONS 

Length of chords, 

C, = 3950-3902-69 = 47-31 ft 

C 2 = C n _, = 50-0 ft 

C n = 5997-09-5950 = 47-09 ft 

Deflection angles (Eq. 11.40) 

206 265 x 50 

for C, = 50 ft, a 2 = — - — = 5156-62" = 1°25'57 



for C, = 47-3 ft, a, = ■ " " = 4879' 



for C n = 47-1 ft, a n = 



2x 1 


L000 


5156-6 


x47-3 


50 




5156-6 


x47-l 



50 



= 4856" 



(say 1°26'00") 

= 1°21 , 19" 
(say 1°21'20") 

= 1°20 , 56" 
(say 1°21'00") 



Check 4879 + 40 x 5156-62 + 4856 = 216 000" 

= 60°00 , 00" 

Example 11.11 Assuming that Example 11.21 is to be set out using 
angular values only, calculate the deflection angle from T, and 7 2 
suitable for use with a 20" theodolite, Fig. 11.29. 

Angles at T may be calculated as follows from the previous 
calculations. 



Fig. 11.29 




a, = 1°21'19" 

+ <x 2 = 1°25'57" 

+ a 3 = 1°25'57" 

+ a 4 = 1°25'57" 

+ a s = 1°25'57" 



Accumulative deflection angle 
say 1°21'20" 

2°47'20" 
4°13'20" 
5°39'20" 
7°05'00" etc. 



CIRCULAR CURVES 585 

Angle at T 2 , by Eq. (11.44): 

j8, = 360 -60 + a, = 301°21'20" 
B 2 = 300 +a,+ a 2 = 302°47'20" 
B 3 = 300 + a, + a 2 + a 3 = 304°13'20" etc. 



Example 11.12 Fig. 11.30 shows the centre lines of existing and 
proposed roadways in an underground shaft siding. It is proposed to 
connect roadways A-B and no. 1 shaft - C respectively by curves 
BD and CD, each 100 ft radius, and to drive a haulage road from D 
in the direction DE on a line tangential to both curves. B and C are 
tangent points of the respective curves and D is a tangent point com- 
mon to both curves. 



N!1 shaft 




SKetch-not to scale 



Fig. 11.30 



Co-ordinates of A N 2311-16 ft, E 2745-98 ft. 
Co-ordinates of no. 1 shaft N 2710-47 ft, E 3052-71 ft. 
Bearing of no. 1 shaft - C 186°30'00". 
Distance of no. 1 shaft — C 355 ft. 
Bearing of roadway A — B 87°23 50 . 

Calculate the distance from A to the tangent point B of the curve 
BD, the co-ordinates of B and the bearing of the proposed roadway 

DE. 

(M.Q.B./S) 



586 



SURVEYING PROBLEMS AND SOLUTIONS 

> N! 1 shaft 



nB7* 23' 50" §_ 






Bearing 177°23'50" 
100ft 




Bearing 096 # 30' 
100ft 



Fig. 11.31 



Construction (Fig. 11.31) Draw AX 100 ft perpendicular to AB Join 
X0 2 ; 0,0 2 . 



Co-ordinates of C 



AE^SSSsine^O' 

N 

11 no. i 

AN,_ c 355cos6°30' 



3052-71 
-40-19 
3012-52 
2710-47 
-352-72 
2357-75 



Co-ordinates of 0, 

E c 3012-52 

AE^lOO sin85°30' +99-36 

E , 3111-88 

N c 2357-75 

AN ^100 cos 85°30' - 11-32 

N a 2346-43 



CIRCULAR CURVES 587 

Co-ordinates of X 

E A 2745-98 

AE4^100sin2 o 36'10" + 4*54 



E x 2750-52 

N^ 2311-16 

N^ 100 cos 2°36 ' 10" -99-90 

Nv 2211-26 



_. 3111-88 - 2750-52 _, 361-36 

Bearing XO. = tan MAt . Af% — tztt~Z7 = tan T^rTZ 

& ' 2346-43 - 2211-26 135-17 

= N 69°29'29" E 

Length XO, = 361-36 cosec69°29'29" 

= 385-81 ft 

In triangle X0,0 2 , 

XO, sin 0, 



XO. = 



sinO 



2 
'r»/\» 



Bearing X0 2 = 87°23'50' 

XO t = 69°29'29" /. Angle X = 17°54'21" 

0, X sin X 385-81 sin 17°54'2l" 
sin0 2 - — jg— - — 

2 = (36°22'33") or 180 - 36°22'33" 
= 143°38'27" 

385-81 sin (180 - 143°38'27" - 17°54'21") 



Thus XO, = 



2 sinl43°38'27" 

385-81 sin 18°27'12" 



sin36°22'33' 
Co-ordinates of B 



= 205-91 ft = AB 



E A 2745-98 

A E AB 205-91 sin 87°23'50" +205-70 

E B 2951-68 

N4 2311-16 

Mi AB 205-91 cos87°23'50" + 9-35 

N fl 2320-51 



588 SURVEYING PROBLEMS AND SOLUTIONS 

Bearing X0 2 = AB = 087°23'50" 
Angle X0 2 0, = 143°38'27" 

231°02'17" 
- 180° 

Bearing 2 0, 051°02'17" 

+ 90° 



Bearing DE 141°02'17" 

Ans. AB 205-91 ft. 

Co-ordinates of B, E 2951-68 N 2320-51. 

Bearing of DE, 141°02'17". 

Exercises 11(c) 

12. AB and CD are straight portions of two converging railways 
which are to be connected by a curve of 1500 ft radius. The point of 
intersection of AB and CD produced is inaccessible. X and Y are 
points on AB and CD respectively which are not intervisible and the 
notes of a theodolite traverse from one to the other are as follows: 



Line 


Horizontal 


Angle 


Distance (ft) 


Xa 


AXa 


260° 10' 


160 


ab 


Xab 


169°00' 


240 


be 


abc 


210°30' 


300 


cY 


bcY 


80°00' 


180 


YC 


cYC 


268°40' 


_ 



Calculate the apex angle and the position of the start and finish 
of the curve relative to X and Y . 

(M.Q.B./S Ans. 91°40'; T,X 1234-42 ft; T 2 Y 780-09 ft) 

13. To locate the exact position of the tangent point T 2 of an exist- 
ing 500 ft radius circular curve in a built-up area, points a and d were 
selected on the straights close to the estimated positions of the two 
tangent points T, and T 2 respectively, and a traverse abed was run 
between them. 

Station Length (ft) Deflection Angle 
a 9°54' R 

b ab 178 19°36' R 

c be 231 30°12' R 

d cd 203 5° 18' R 

The angles at a and d were relative to the straights. Find the distance 
T 2 d. (L.U. Ans. 34- lft) 



2° 36-6' 


72 - 


4° 06-6' 


7° 06-6' 


75 - 


8° 36-6' 


l°36-6' 


T.P. 


13° 00') 



CIRCULAR CURVES 589 

14. Two straights AI and BI meet at / on the far side of a river. 
On the near side of the river, a point E was selected on the straight 
AI, and a point F on the straight BI, and the distance from £ to F 
measured and found to be 3*40 chains. 

The angle, AEF, was found to be 165°36' and the angle BFE 
168°44' . If the radius of a circular curve joining the straights is 20 
chains, calculate the distance along the straights from E and F to the 
tangent points. 

(I.C.E. Ans. 3-005 ch; 2-604 ch) 

15. The centre line of a proposed railway consists of two straights 
joined by a 3° curve. The angle of deflection between the straights is 
26° and the chainage (increasing from left to right) of their intersection 
is 7367 ft. 

Calculate the deflection angles from the tangent for setting out the 
circular curve from the first tangent point by pegs at every 100 ft chain- 
age and check on to the second tangent point. 
(I.C.E. Ans. peg 70- l°06-6' 71- 
73 - 5° 36-6' 74 - 
76 - 10° 06-6' 77 - 

16. Outline three different methods for setting out a circular curve of 
several hundred feet radius using a chain and tape and without using a 
theodolite. Sketch a diagram for each method and quote any formulae 
used in the calculations associated with each method. 

The centre line of a certain length of a proposed road consists of 
two straights with a deflection angle of 30°00' and joined by a circular 
curve of 1000 ft radius; the chainage of the tangent point on the first 
straight is 3630 ft. The curve is to set out by deflection angles from 
this tangent point using a theodolite which reads to 20", and pegs are 
required at every 100 ft of through chainage and at the second tangent 
point. 

Calculate these deflection angles and any other data that could be 
used in the field for checking the position of the second tangent point. 

(I.C.E. Ans. 2°00'20"; 4°52'20"; 7°44'00" 
10°36'00"; 13°28'00"; 15°00'00") 

17. The tangent length of a simple curve was 663-14 ft and the de- 
flection angle for a 100 ft chord 2° 18' . 

Calculate the radius, the total deflection angle, the length of the 
curve and the final deflection angle. 

(L.U. Ans. 1245-5 ft; 56°03'50"; 1218-7 ft; 28°01'55") 

18. (a) What is meant by the term 'degree of curve' (D)? 

State the advantages, if any, of defining a circular curve in this 
way. Show how the degree of curve is related to the radius of the curve 
(R). 



590 SURVEYING PROBLEMS AND SOLUTIONS 

(b) Two straights EF and FG of a proposed road intersect at 
point F. The bearings of the straights are: 
EF 76°12' 
FG 139°26' 
The chainage of E is 11376-0 (113 + 76-0) and the distance EF 
is 2837-6(28 + 37-6). 

Calculate the chainages of the tangent points and prepare a table 
of the deflection angles from the tangent point on GF for pegs at whole 
100 ft chainages for a 5° curve connecting the two straights. 

Explain with the aid of a diagram how you would set out this curve. 
(R.I.C.S./L Ans. R = 1146 ft; 10670-3; 11935-4) 

19. A circular curve XT of 1000 ft radius joins two straights AB and 
BC which have bearings of 195° 10' and 225°40' respectively. At what 
chainage from X measured along the curve, will the curve be nearest 
to point B? 

If this point of nearest approach to B be point W what is the 
bearing of WB? 

(R.I.C.S./G Ans. 266-2 ft; 120°25') 

20. (a) The lines AB and BC are to be joined by a circular curve 
of radius 3000 ft. The point B of intersection of the lines is inaccess- 
ible. 

The following values have been measured on the ground: 
angle AMN = 146°05' , angle MNC = 149°12' , MN = 2761ft; 

and the chainage of M is 25342ft (measurement from A). 

Calculate the length of the curve and the chainage of the beginn- 
ing and end of the curve. 

(b) Deduce formulae for setting out intermediate points on the 
curve by using steel band, linen tape, optical square and ranging rods 
only. 

(R.I.C.S./L Ans. 3388-6; 25004-7; 28393-3) 

21. A BC D is a plot of land, being part of a block. 

It is required to round off the corner by a circular curve tangential 
to the boundaries at B and C. What is the radius of the curve to the 
nearest tenth of a foot? 

Angle BAD = 90° AB = 100 ft 

Angle ADC = 52°38' AD = 140 ft 

(R.I.C.S./L Ans. R = 28-2 ft) 

22. A railway boundary CD in the form of a circular arc is intersect- 
ed by a farm boundary BA in E. Calculate this point of intersection 
E. 

The radius of the curve is 500 ft and the co-ordinates in feet are: 



A 


E 7525-7, 


B 


E 7813-4 


C 


E 8009-3 


D 


E 7101-2 



CIRCULAR CURVES 591 

N 21951-7 
N 20 163-3 
N 21 179-6 
N 21074-3 

(Ans. E 7610-5 N 21424-7) 

23. A 750 ft length of straight connects two circular curves which 
both deflect right. The first is of radius 1000 ft, the second is of 
radius 800 ft and deflection angle 27° 35' . The combined curve is 
to be replaced by a single circular curve between the same tangent 
points. 

Find the radius of this curve and the deflection angle of the first 
curve. 

(L.U. Ans. R = 1666-5 ft; 31°33') 

24. In a level seam two roadways AB and DC are connected by 
roadways AD and BC. Point B is 820 ft due East of A, D is 122 ft 
due North of A, and C is 264 ft due North of B. It is proposed to 
drive a circular curve connecting BA and DC and tangential to BA, 
AD and DC. Calculate the radius of the curve and the distances from 
A and D to the tangent point of the curve on the lines AB and DC 
respectively. 

(M.Q.B./S Ans. 66'24 ft; 66'24 ft; 55'76 ft) 

25. The co-ordinates of two points A and B are: 

E N 

A 0-0 399-60 

B 998-40 201-40 

A straight line AC bears 110°30' and intersects at C a straight 
line BC bearing 275° 50'. The chainage of A is 2671-62 ft. Calculate 
the lengths of AC and CB. 

The two straights are to be joined by a curve of 500 ft radius. Cal- 
culate the chainage of the tangent points and of B. 

(L.U. Ans. 377-97ft; 647'71ft; 2985'24ft; 3113'23ft; 3696'59ft) 

11.8 Compound Curves 

Compound curves consist of two or more consecutive circular arcs 
of different radii, having their centres on the same side of the curve. 

There are seven components in a compound curve made up of two 
arcs: 



Two radii, /?, and R. 

Two tangent lengths, t x and t 2 

An angle subtended by each arc, a 1 and a 2 . 



592 



SURVEYING PROBLEMS AND SOLUTIONS 



A deviation angle at the intersection point /, = a,+ a. 
At least 4 values must be known. 




Fig. 11.32 
In Fig. 11.32, 

AB sin a ^' tan \ a ' + R * tan 1 a ^ sin a 2 



A/ = 



sin <f> 
TJ = T,A + AI = f, 



sin<£ 



* , = /? , tan -= a , + 



(#, tan 2 a, + i? 2 tan-2<x 2 ) sin a. 



sin <fi 



t , sin <f> = R y tan -1 a, sin <£ + tf, tan -^ a, sin a 2 + R 2 tan -1 a 2 sin a 2 

sin 2 o^ 2 sin ^ a 2 cos ^ a 2 



= R , tan -_- a, (sin <f> + sin a 2 ) + /?, 



cos - a 2 



= R, 



sini(<£- a 2 )2sin^(0+ a 2 ) cos -!(<£- a 2 ) 
cos 4(0- a 2 ) 



+ 2&, sin 2 \ a 



= /?,( c °s a 2 - cos 0) + /? 2 (l-cosa 2 ) 

= fl,{(l - COS<£)-(l - COSOj) } + /? 2 (1 - COSCL,) 

= (R 2 - R, )(1 - cos o^) + R, (1 - cos 0) 
^sincft = (/? 2 - /?,) versinea 2 + /^versineqS (11.46) 

Similarly, it may be shown that 
f 2 sin<£ = (/?,- R z ) versinea, + /? 2 versine<£ (11.47) 



CIRCULAR CURVES 



593 



An alternative solution is shown involving more construction but 
less trigonometry, Fig. 11.33. 




Construction 



Fig. 11.33 

Produce arc T,T 3 of radius R, to 8 
Draw 0,3 parallel to 2 T 2 

BC parallel to IT 2 

BD parallel to T 3 2 

T, A perpendicular to 0, B 

T, E perpendicular to T 2 1 (produced) 



N.B. (a) T 3 BT 2 is a straight line. 

(b) T 3 0,0 2 is a straight line. 

(c) BD = 0,0 2 = DT 2 = R 2 -R,. 
T,E = AB + CT 2 

= (0,8-0,4) + (DT 2 -DC) 



i.e. 



*, sin<£ = /?,-£, cos(a, + a 2 ) + \(R 2 - #i) - (K 2 - #i) cos a 2 \ 

= #,(1 - cos 96) + (R 2 - #i)(l - cosa 2 ) 
f, sin<£ = R, versinec6 + (R z -R x ) versine a 2 (11.48) 



By similar construction, Fig. 11.34, 
FT 2 = GH - HJ 

= (0 2 H - 2 G) - (HK - JK) 



t 2 sincf> = R 2 vers (f> + (/?,_ R 2 ) versa, 



(11.49) 



594 



SURVEYING PROBLEMS AND SOLUTIONS 




Fig. 11.34 

Example 11.13. Given K, = 20 m, R 2 = 40 m, T,/ = 20-5 m, 
<f> = 80° 30', it is required to find length T 2 l. 
From Eq. (11.48), 

T,/ sin<£ - /?, versine^ 



versine a, = 



R 2 - R t 
20-5 sin 80° 30' - 20 vers 80° 30' 



40 - 20 



3-520 
20 



then 

From Eq. (11.49), 



a 2 = 34° 31' 

a, = 45° 59' i.e. <f> - a z 

T , _ R 2 vers<ft - (R 2 - R } ) vers a i 
sin <p 

40 vers80°30' - 20 vers 45° 59' 



sin 80° 30' 



27-67 m 



Alternative solution from first principles (Fig. 11.35) 
Construction 

Join 0,/ 

Draw 0,P parallel to 2 T 2 
0,Q parallel to IT 2 



CIRCULAR CURVES 



595 



In triangle TJO,, 
6 = tan-'^A 



* -t 20 
— = tan ' 

7,/ 20-5 




Fig. 11.35 



= 44° 18' 
j8 = 180 - (0 + <f>) = 55° 12' 
In triangle IPO, , 
IP = 0,1 cos j8 

= R, cosec cos/3 

= 20 cosec 44° 18' cos 55° 12' 

= 16-35 

0,P = 0,1 sinj8 

= 20 cosec 44° 18' sin 55° 12' = 23-52 

2 Q = 2 T 2 - QT 2 = R 2 - 0,P 

= 40 - 23-52 

= 16-48 

In triangle 0,0 2 Q, 

a 2 =cos-M = cos-1^48 = 3^ 
2 0,0, 20 

••• a, = 45° 59' 
0,Q = 0,0 2 sina 2 = 20sin34°3l' = 11-32 m 
IT 2 = IP + PT 2 = IP + 0,Q = 16-35 + 11-32, 

= 27-67 m 



Example 11.14. AB and DC are the centre lines of two straight 
portions of a railway which are to be connected by means of a com- 
pound curve BEC, BE is one circular curve and EC the other. The 
radius of the circular curve BE is 400 ft. 

Given the co-ordinates in ft : B N400 E200, C N593 E 536, and 
the directions of AB and DC NE25°30' and NW76°30' respectively, 
Calculate (a) the co-ordinates of E, (b) the radius of the circular curve 
EC. (R.I.C.S./M ) 

Formulae are not very suitable and the method below shows an 
alternative from first principles. 

Bearing BC = tan"' 536 ~ 200 = 60°07' 
593 - 400 

BC = 193 sec 60° 07' 

<f> = 180 - 76° 30' - 25° 30' = 78° 00' 



596 



SURVEYING PROBLEMS AND SOLUTIONS 



E 53§i_— — ---s»*^ 




Fig. 11.36 

In triangle S/,C, 

B = 60° 07' - 25° 30' = 34° 37' 
C = 78° 00' - 34° 37' = 43° 23' 
8/, = BC sine cosecc6 

= 193 sec 60° 07' sin 43° 23' cosec 78° = 272-0 ft 
C/, = BC sin B cosec ^ = 225-0 ft 



Bh = 400 tan Z§ 
2 2 



/ 2 G = 323-9 ft 



/,/ 2 = 323-9 - 272 = 51-9 ft 

I,J = 51-9 cos 78° = 10-8 ft 

IgJ = 51-9 sin 78° = 50-8 ft 

FG = CH = 2 L = / 2 G + /,./- /,C = 323-9 + 10-8 - 225-0 

= 109-7 



In triangle CGH, 



a 2 
2 



22. = tan- 1 GH/CH 



a 2 = 2 tan"' I 2 J/CH = 2 tan"' 

= 2 x 24° 50' = 49° 40' 
a, = 78°- 49° 40' = 28° 20' 



50-8 
109-7 



CIRCULAR CURVES 597 

n triangle 0^0 2 L, 

0^0 2 = 2 L coseca 2 

= 109-7 cosec49°40' = 143-9 

.*. R 2 = 400 - 143-9 = 256-1 

to find the co-ordinates of E, 

Bearing BE = 25° 30' + 14° 10' = 39° 40' 
BE = 2 x 400 sin 14° 10' 

Partial Lat. = BE cos 39° 40' = 150-71 

Total Lat. = N 550-7 

Partial Dep. = BE sin 39° 40' = 124-98 

Total Dep. = E 325-0 

Iheck 

From Eq. (11.48), 

/, sin<£ = 272 sin 78° = 266-06 ft 
R, versine^ = 400(1 - cos 75°) = 316-83 ft 
(R 2 - R,) versinea 2 = (256-1 - 400) (1 - cos49°40') = -50-76 ft 
.*. /?, versc6 + (R 2 - /?,) vers a 2 = 316-83 - 50-76 = 266-07 ft 



Example 11.15. Undernoted are the co-ordinates in ft of points on the 
■espective centre lines of two railway tracks ABC and DE 



Co-ordinates . 


A 










B 


E 525-32 


N 52-82 




C 


E 827-75 


N 247-29 




D 


E 10-89 


S 108-28 




E 


E 733-23 


S 35-65 



The lines AB and DE are straight and B and C are tangent 
>oints joined by circular curve. It is proposed to connect the two 
racks at C by a circular curve starting at X on the line DE, C being 
t tangent point common to both curves. Calculate the radius in chains 
>f each curve, the distance DX and the co-ordinates of X. 

(M.Q.B./S) 
In Fig. 11.37, 

Bearing AB = tan" 1 + 525-32/+ 52*82 
= N 84° 15' 30" E 



598 

Bearing DE = tan 



SURVEYING PROBLEMS AND SOLUTIONS 

_, 733-23 - 10-89 



-35-65+ 108-28 

= N 84° 15' 30" E 

As bearings are alike, C, B and X 
must lie on the same straight line. 

Bearing BC = bearing XC 

= tan-1 827 ' 75 - 525-32 
247-29 - 52-82 

= N 57° 15' 29" E 

Length BC = AN sec bearing 

= (247-29 -52-82) sec 57°15'29" 

= 359-56 

Angle (a) subtended at centre = 2 x angle BXE 

= 2 x (84° 15' 30" - 57° 15' 29") 




Fig. 11.37 



= 54°00'02' 



Radius 2 C = ±BC cosecia 



= I x 359-56 cosec27°00'01" 
2 



396 ft = 6 chains 



Bearing DB = tan-' 525 ' 32 



10-89 



= N 72° 36' 41" E 



52-82 + 108-28 
Length DB = 161-10 sec 72° 36' 41" = 539-06 
Bearing DE = 084° 15' 30" 

DB = 072° 36' 41" .-. Angle BOX = 11°38'49" 

BC = 057° 15' 29" /. Angle DBX = 15° 21' 12" 

In triangle DBX, 



DX = 
BX = 



BD sing _ 539-06 sin 15° 21' 12" 

sinX ~ ■ sin 27° 00' 01" 
BD sin D 539-06 sin 11° 38' 49" 



sinX sin 27° 00 '01" 

Corordinates of X 

Line DX N 84° 15' 30" 314-38 ft 

.'. AE = 314-38 sin 84° 15' 30" = +312-80 

E D = 10-89 

E^ = 323-69 

AN = 314-38 cos 84° 15' 30" = + 31-45 

Nn = -108-28 



= 314-38 ft 
= 239-71 ft 



CIRCULAR CURVES 599 

N x = -76-83 
CX = BX + BC 

= 239-71 + 359-56 = 599-27 ft 
Radius 0,C = |CXcosec27°00'0l" 

= 1 x 599-27 cosec27°00'0l" 

= 659-986 ft 
a* 660-0 ft = 10 chains 
Ans. O z C = 6 chains radius 
O^C = 10 chains radius 
DX = 314-38 ft 
Co-ordinates of X E 323-69, S 76-83 

Exercises 11(d) (Compound curves) 

26. A main haulage road AD bearing due north and a branch road DB 
bearing N87°E are to be connected by a compound curve formed by 
two circular curves of different radii in immediate succession. The 
first curve of 200 ft radius starts from a tangent point A, 160 ft due 
South of D and is succeeded by a curve of 100 ft radius which termin- 
ates at a tangent point on the branch road DB. 

Draw a plan of the roadways and the connecting curve to the scale 
1/500 and show clearly all construction lines. 

Thereafter calculate the distance along the branch road from D to 
the tangent point of the second curve and the distance from A along 
the line of the first curve to the tangent point common to both curves. 

(M.Q.B./S Ans. 120-2 ft; 145-4 ft) 

27. The intersection point / between two railway straights T'XI and 
IYT is inaccessible. Accordingly, two arbitrary points X and Y are 
selected in the straights and the following information is obtained: 

Line Whole circle bearing 

T'XI 49° 25' 

IYT" 108° 40' Length XY = 1684-0 ft 

XY 76° 31' ChainageofX = 8562-3 ft 

The straights are to be connected by a compound circular curve 
such that the arc T'C, of radius 3000 ft, is equal in length to the arc 
CT", of radius 2000 ft, C being the point of compound curvature. Make 
the necessary calculations to enable the points T', C and T" to be 
pegged initially, and show this information on a carefully dimensional 
sketch. 



600 



SURVEYING PROBLEMS AND SOLUTIONS 



Determine also the through chainages of these points. 

(L.U. T'l = 1489-1 ft chainage T 8115*9 ft 
T'l = 1235-5 ft chainage T" 10 597-7 ft 

C 9356-8 ft) 



11.9 Reverse Curves 

There are four cases to consider: 

(1) Tangents parallel (a) Radii equal (b) Radii unequal. 

(2) Tangents not parallel (a) Radii equal (b) Radii unequal. 
Tangents parallel (Cross-overs) (Fig. 11.38) 

T, T 2 , 0, 2 and /, / 2 are all straight lines intersecting at E, the 
common tangent point, 



Fig. 11.38 




In Fig. 11.38, 



Angle BT^E = T, £/, = ^ 0, 



T EI = —T O E 



= \bIJ 2 = a 

= icy, = a 



Angle CT 2 £ 

Triangle T,/,fi is similar to triangle ET 2 / 2 

Triangle 1,^0, is similar to triangle E0 2 T 2 

Tangents not parallel (Fig. 11.39) 
N.B. T, T 2 does not cut 0^0 2 at E. 



and 




Fig. 11.39 



CIRCULAR CURVES 



601 



N.B. These solutions are intended only as a guide to possible 
methods of approach. 
Tangents parallel (Fig. 11.40) 

Bisect T,£ and T 2 E. Draw perpendiculars PO^ and QQ 2 . 



Fig. 11.40 



*i 






\-» / ! 

NX j 

— * — ^ 



T,E = 2T,P = 2K, sina 
T 2 E = 2T 2 Q = 2K 2 sina = T,T 2 - T,£ 
2R Z sin a = I, T 2 - 2K, sin a 
XT, 



If R, = R z , 



R, = 



R = 



2 sina 



T T 

4 sina 



- *, 



(11.50) 



(11.51) 



Example 11.16 Two parallel railway lines are to be connected by a 
reverse curve, each section having the same radius. If the centre lines 
are 30 ft apart and the distance between the tangent points is 120 ft, 



what will be the radius of cross-over? 



In Fig. 11.41, 

T,A = 30 ft 



T X T = 120 ft 




In triangle T x T Z A, 




T.A 
sina = —— - = 
i, i z 


30 
120 




In triangle T x PO x , R = T, 0, = 



T, P T, T 2 



Fig. 11.41 

120 



sina 4 sina 4x30/120 



= 120 ft 



Tangents not parallel, radii equal (Fig. 11.42) 
Construction 

Draw 0, S parallel to T, T 2 , PO x perpendicular to T, T 2 , Q per- 
pendicular to T, T 2 



602 



SURVEYING PROBLEMS AND SOLUTIONS 



In triangle T, PO x , 

I, P = R sin a, 

PO, = R cos a, = QS 

In triangle 2 T 2 Q, 

T 2 Q = R sina 2 

2 Q = R cosa 2 

2 S = 2 Q + QS 

= R cosa 2 + R cos a, 
= i?(cosa, + cosa 2 ) 




2 S 



P ' sin 0^ 2 



Fig. 11.42 



= sin 



_ l /?(cosa, + cosa 2 ) 



2R 



0,S 



= 2R cos jS 

= T,T 2 -(T,P + QT 2 ) 



= T y T 2 - K(sina, + sina 2 ) 
T T 
2 cos|3 + sina, + sina 2 



(11.52) 



Example 11.17 Two underground roadways AB and CD are to be 
connected by a reverse curve of common radii with tangent points at 
B and C. If the bearings of the roadways are AB S 83° 15' E and CD 
CD S 74° 30' E and the co-ordinates of B E 1125*66 ft N 1491-28 ft, 
C E 2401-37 ft N 650*84 ft, calculate the radius of the curve. 

(M.Q.B./S) 




Fig. 11.43 



The bearings of the tangents are different and thus the line BC 
does not intersect with 0»0 2 at the common tangent point although at 
this scale the plotting might suggest this. 

In Fig. 11.43, 



CIRCULAR CURVES 



603 



n . Qn , -.2401-37-1125-66 
Bearing BC = tan ' —T^z—rz — « AM nn 



= tan" 



650-84-1491-28 
, 1275-71 



-840-44 
= S 56° 37' 23" E 

Length BC = 840-44 sec 56° 37' 23" = 1527-67 ft 

Bearing BC S 56° 37' 23" E 

AB S 83° 15' 00" E .-. a, = 26° 37' 37" 

CD S 74° 30' 00" E ctj, = 17° 52' 37" 

/8 = sin" 1 I (cos 26° 37' 37" + cos 17° 52' 37") 
= 67° 20' 39" 

0,S = 2R cos 67° 20' 39" 
BP = R sin a, 

= R sin 26° 37' 37" 
QC = R sina 2 

= R sin 17° 52' 37" 
BC = BP + 0,S + QC 
.-. 1527-67 = R sin 26° 37' 37" + 2R cos 67° 20' 39" + R sin 17° 52' 37" 
By Eq. (11. 50), 

R 1527-67 

~ sin 26° 37' 37" + 2 cos 67° 20' 39" + sin 17° 52' 37" 
= 1001-31 ft 

Tangents not parallel, radii not equal (Fig. 11.44) 




Fig. 11.44 

Construction 
Join 0, T 2 . Draw 0, P perpendicular to T, T 2 

T, P = /?, sin a, 
O^P = #, cos a, 



604 SURVEYING PROBLEMS AND SOLUTIONS 

PT 2 = T y T 2 - T,P = T,T 2 - /?, sina, 

= tan '-^=- = tan (11.53) 

Pi 2 7, T 2 - /<! sin a, 

n t °' P _ R i cosa i 

sintf ~ sin0 

In triangle 0, 2 I 2 , 

0,01 = OJl + 0,T 2 2 - 20 2 T 2 0J 2 cos 0,f 2 2 

R* cos 2 <x, 



(fl, + R 2 y = K 2 2 + 



sin 2 

2/? 2 K, cos a, cos {90 - (a 2 - 0)1 
sin0 

2 22 #, 2 cos 2 a,- 2#,i? 2 cosa, sin(a 2 -0)sin0 

/v. + 2R.R? + Rf = /v., H ; — 2 ~: 

sin 6 
R,(R, cos 2 a, - 2R 2 cos a, sin(a 2 - 0) sin 0) 

/?,(*, + 2K 2 ) = ~i^ 1 2 -—rr ~ 

sin u 

2R 2 sin {sin + cos a, sin(a 2 - 0)} = R,(cos 2 a, -sin 2 0) 

R = /? 1 (cos 2 a 1 - sin 2 0) 

2 2 sin {sin + cosa,sin(a> - 0)} 

Example 11.18 Two straights AB and CD ate to be joined by a cir- 
cular reverse curve with an initial radius of 200 ft, commencing at B. 

From the co-ordinates given below, calculate the radius of the 
second curve which joins the first and terminates at C. 

E N 

Co-ordinates (ft) A 103-61 204-82 

B 248-86 422-62 

C 866-34 406-61 

D 801-63 141-88 

. (R.I.C.S./M) 



A 
103-61 El 
204-82 N 



Fig. 11.45 



CIRCULAR CURVES 605 

In Fig. 11.45, 

Bearing AB = tan- g^*|^j|^ = N 3 3°42'00" E = 033° 42*00' 



i r\r\H 



DC = tan-' ^'^-^l'S - N 13°44'00" E = 013°44'00 
(406*61 -141-88) 



BC = tan"' (866 ' 34 ~ 248 ' 86) = S 88°30'50" E = 091° 29' 10" 
(406-61-422-62) 

Length BC = 617-48/cos88°30'50" = 617-69 ft (T, T 2 ) 

Angle a, = 91° 29' 10" - 033° 42' 00" = 57° 47' 10" 
a = 91 o 29'10"-013°44'00" =77° 45' 10" 

2 

By Eq. (11.53), 

R,cosa, , 200 cos 57° 47' 10" 
v = tan '-srs s ^— = tan „«„-.,„ nnn ... fw0 A n> m» 



TJ 2 - R, sina, " 617-48 - 200 sin 57° 47' 10 

= 13° 22' 20" 

By Eq. (11.54), 

200 (cos 2 57° 47' 10" - sin 2 13° 22' 20") 

2 ~ 2 sin 13° 22' 20" {sin 13° 22' 20" + cos 57° 47' 10" sin(77°45'10" - 13° 22' 20 

= 140-0 ft 

Exercises 11(e) (Reverse curves) 

28. A reverse curve is to start at a point A and end at C with a 
change of curvature at B. The chord lengths AB and BC are respec- 
tively 661-54 ft and 725*76 ft and the radius likewise 1200 and 
1500 ft. 

Due to irregular ground the curves are to be set out using two 
theodolites and no tape and chain. 

Calculate the data for setting out and describe the procedure in 
the field. 

(L.U. Ans. Total deflection angles, 16°; 14°; 
Setting out by 1° 11' 37" deflections) 

29. Two roadways AB and CD are to be connected by a reverse 
curve of common radius, commencing at B and C. 

The co-ordinates of the stations are as follows: 

A 21642-87 m E 37 160*36 m N 

B 21 672*84 m E 37 241*62 m N 

C 21 951*63 mE 37350*44 m N 

If the bearing of the roadway CD is N 20° 14' 41" E, calculate the 

radius of the curve. 

(Ans. 100*0 m) 



606 SURVEYING PROBLEMS AND SOLUTIONS 

30. CD is a straight connecting two curves AC and DB. The curve 
AC touches the lines AM and CD; the curve DB touches the lines 
CD and BN. 

Given: Bearing MA = 165° 13' 

BN = 135° 20' 
Radius of curve AC = 750 m 
DB = 1200 m 
Co-ordinates A E + 1262*5 m N - 1200-0 m 

B0 

Calculate the co-ordinates of C and D. 
(Ans. C E + 1562-54 m N - 1358*58 m; D E + 57-32 m N - 53-09 m) 

31. Two parallel lines which are 780 m apart are to be joined by a 
reverse curve ABC which deflects to the right by an angle of 20° 
from the first straight. 

If the radius of the first arc AB is 1400 m and the chainage of A 
is 2340 m, calculate the radius of the second arc and the chainages of 
B and C. (Ans. 1934 m; 2828 m; 3503*8 m) 

32. Two straight railway tracks 300 ft apart between centre lines 
and bearing N 12° E are to be connected by a reverse or 'S* curve, 
starting from the tangent point A on the centre line of the westerly 
track and turning in a north-easterly direction to join the easterly track 
at the tangent point C. The first curve AB has a radius of 400 ft and 
the second BC has a radius of 270 ft. The tangent point common to 
both curves is at B. 

Calculate (a) the co-ordinates of B and C relative to the zero 
origin at A (b) the lengths of the curves AB and BC. 

(M.Q.B./S Ans. (a) B E 244-53 N 288*99 ft 
C E 409*58 N 484*05 ft 
(b) 394*30 ft; 266*15 ft) 

Bibliography 

THOMAS, W.N., Surveying (Edward Arnold) 

CLARK, D., Plane and Geodetic Surveying (Constable) 

DAVIES R.E., FOOTE F.S., and KELLY, j.w., Surveying Theory and 

Practice (McGraw-Hill) 
SANDOVER, J. A., Plane Surveying (Edward Arnold) 
BANNISTER, A., and RAYMOND, S., Surveying (Pitman) 
JAMESON, A.H., Advanced Surveying (Pitman) 
SALMON, V.G., Practical Surveying and Fieldwork (Griffin) 
JENKINS, R.B.M., Curve Surveying (Macmillan) 
KISSAM, P., Surveying for Civil Engineers (McGraw-Hill) 
SEARLES, W.H. and IVES, H.C., rev. by P. KISSAM, Field Engineering, 

22nd Ed. (John Wiley) 
HIGGINS, A.L., Higher Surveying (Macmillan) 



12 VERTICAL AND TRANSITION CURVES 
12.1 Vertical Curves 

Where it is required to smooth out a change of gradient some form 
of parabolic curve is used. There are two general forms, (a) convex or 
summit curves, (b) concave, i.e. sag or valley curves. 

The properties required are: 

(a) Good riding qualities, i.e. a constant change of gradient and a 
uniform rate of increase of centrifugal force. 

(b) Adequate sighting over summits or in underpasses. 

The simple parabola is normally used because of its simplicity 
and constant change of gradient, but recently the cubic parabola has 
come into use, particularly for valley intersections. It has the advant- 
age of a uniform rate of increase of centrifugal force and less filling is 
required. 

Gradients are generally expressed as 1 in x, i.e. 1 vertical to x 
horizontal. For vertical curve calculations % gradients are used : 



e.g. 



linx = m% 

x 
1 in 5 = 20% 



Gradients rising left to right are considered +ve. 

Gradients falling left to right are considered - ve. 

The grade angle is usually considered to be the deflection angle = 
difference in % grade. By the convention used later, the value is given 
as q%-p%, Fig. 12.1. 





-Q -p = -(q+p)v. 




q-(-p)**(q+p)»l. 



608 



SURVEYING PROBLEMS AND SOLUTIONS 



12.2 Properties of the Simple Parabola 

y = ax 2 + bx + c 

& = lax + b 
dx 



2a 



for 



max or mm 



(12.1) 



(12.2) 



dJz 



, = 2a, i.e. constantrateofchangeofgradient (12.3) 

If a is +ve, a valley curve is produced. 

If a is -ve, a summit curve is produced. 

The value of b determines the maximum or minimum position along the 
x axis. 

The value of c determines where the curve cuts the y axis. 

The difference in elevation between a vertical curve and a tangent 
to it is equal to half the rate of change of the gradient x the square of 
the horizontal distance from the point of tangency, Fig. 12.2. 




Fig. 12.2 



If 



y = ax 2 + bx + c, 

-¥- = 2 ax + b (grade of tangent) 
dx 



When x = 0, grade of tangent = +b 
value of y = + c 
In Fig. 12.2, if the grade of the tangent at x = is b, 



VERTICAL AND TRANSITION CURVES 



609 



QR = b x x i.e. + bx 
RS = +c 

PQ = ax 2 

as y = PQ + QR + RS 

= ax 2 + bx + c 

The horizontal lengths of any two tangents from a point to a verti- 
cal curve are equal, Fig. 12.3. 

PB = ax? = ax\ (1) 

. . x x — x z 

i.e. the point of intersection B of two gradients is horizontally mid- 
way between the tangent points A and C. 

A chord to a vertical curve has a rate of grade equal to the tangent 
at a point horizontally midway between the points of intercept, Fig. 12.3. 



From (1), 

AA X = CC X = ax 2 
AA, is parallel to CC, 
ACC^A % is a parallelogram 
AC is parallel to i4,C,. 



Fig. 12.3 

12.3 Properties of the Vertical Curve 

From the equation of the curve, y = ax 2 + bx + c as seen previ- 
ously. 

















C 


A 








P^ 






c, 


*i 




















*i 




B 


*a 
















Fig. 12.4 



610 SURVEYING PROBLEMS AND SOLUTIONS 

In Fig. 12.4, 

AB = ax 2 

BC = bx 
(N.B. For summit curves, a is negative) 

.'. the level on the curve at A = ax 2 + bx 
DT 2 = al z = ET Z - ED 

= JL - JpL = Kg-p) 

200 200 200 

Half the rate of change of gradient 
(N.B. a will be negative when p > q) 

a = g-P 
200/ 
Length of curve 



(12.4) 



' = 4^ ( 12 - 5 ) 

200 a 

It is common practice to express the rate of change of gradient 
2a as a % per 100 ft. 

.'. the horizontal length of curve may be expressed as 

100 x grade angle 



/ 



% rate of change of grade per 100 ft 
100 (q - p) 



2a%per 100 
Distance from the intersection point to the curve 
IF = KT, = FG = JT 2 

2 



(12.6) 



■ •©" 



L. x q ~ p = l ( q ~ p ) (12.7) 

4 200/ 800 



Maximum or minimum height on the curve 
H = ax 2 + bx 

= (q-p)x 2 + p±_ 

200 / 100 

for max or min 

dH = (q - p) x | p = Q 

dx 100/ 100 

then x = ~ p/ = pl (12.8) 

<?-P P- Q 



VERTICAL AND TRANSITION CURVES 



611 



Example 12. 1 A parabolic vertical curve of length 300 ft is formed at 
a summit between grades of 0*7 per cent up and 0*8 per cent down. 
The length of the curve is to be increased to 400 ft, retaining as much 
as possible of the original curve and adjusting the gradients on both 
sides to be equal. Determine the gradient. (L.U.) 



m. iwiw <-'v£. y.».*..-ry, 


a 


_ q - p 

200/ 










-0-8 - 


- 0-7 


1-5 






" 200 x 


300 


60000 


If the gradients are to be made e 


qual, 


p = q. 


P 


+ Q 


= 200 a/ 








2p 


= 200 x 


400 


1-5 
60000 




P 


= 1% 








12.4 


Sight Distances (s) 



12.41 Sight distances for summits 
(1) s > / 
In Fig. 12.5, 
Let / = horizontal length of vertical curve 
sight distance 

AC = height of eye above road at A 
OL = height of object above road at 
distance of vehicle from tangent point T, 
distance of object from tangent point T 2 

TJ = JP- + 12- = -L (n + q) 
2 200 200 200 KP qJ 



I 

s 

A, 

<*, 



(1)s>< 




Fig. 12.5 Principles of sight distances 



612 



AB 



Similarly, 







SURVEYING PROBLEMS AND SOLUTIONS 




. T 2 J p + q 
tana = ■==-— = nnn 
T,J 200 




*£ d,tana = d 'P *<? + «> _ d ' (2p p- 
100 ' 100 200 200 


-9) 


= J±(p-q) 
200 





MO - d tan a - hi - dz ^P + g) £^ _ A. / n , . 2 /rt 
MU - d 2 tana - — _ 200 100 ~ 200 {p + q ~ 2q) 



200 



ft, = AB + BC 

= Jl-(p-q) + a(i) 2 
200 V2/ 



but a is negative, i.e. a = ^ — Ei 

200/ 

*' - 200 tP *' 200/ X 4 



- A. r D - a) + /( P~^ - P ~ g T4rf + /I 
~ lOO tP ^ + ^00 8O0T L ' + J 



Similarly, 



h 2 = OM + BC = J=LjLi[4d a+ Z] 

, _ . . 800ft, - /(p-g) 800ft 2 - /(p-g) 

_ _ d} + dz = __ + __ 

400(6, +h 2 ) - Up-q) 
2(p - q) 

2s -21 = 400 ^ + *a) _ , 
P - q 

I = 2s - 4 00(^i +^2) 
P - q 



(12.9) 



If s = /, 



/ = 2/ - 4 00(^i + *fc) 
P - 9 

/ = 400(ft« + ^) (1210) 

P - 9 



(2) s < / 

In Fig. 12.6, ft, = ad] .-. d. = )!h 



h, = adt 



VERTICAL AND TRANSITION CURVES 

yjh 2 



613 



*. dp = 



y/a 



V" 



(2) S<1 



But p - a 

a = — 

200/ 

200/ 



tV^i +\f h z]' 



BRSiSTO! 



Fig.12.6 



/ = 



p - q 

S 2 (p - q) 



N.B. If h x = h 2 = h: 
(i) s > I 



(12.11) 



(ii) s = / 
(iii) s < I 



I = 2s - 
= 2s - 
I = 

/ = 



400 x 2ft 
p - q 
800 ft 



p - q 



800/2 
p - q 

s\p - q) 



200.[2(V/i)] 2 
s 2 (p - g) 



800 h 



(12.12) 
(12.13) 

(12.14) 



12.42 Sight distances for valley curves 

Underpasses 

Given: clearance height H, 

height of driver's eye above road /i,, 
height of object above road h 2 , 
sight length s, 
gradients p% and q%. 

(1) s > I 

Let the depth of the curve below the centre of the chord AD (the 
distance between the observer and the object) be M, Fig. 12.7. 

GL _ DE 

s 



tana = 



GL 



is 

GJ + JI + IL 



4 200 



By Eq. (12.4), 



a = 



_ q - P 
200/ 



614 



SURVEYING PROBLEMS AND SOLUTIONS 




itiie&ft 



Fig. 12.7 Sight distances for valley curves 
and from above, 

DE = 2GL 



Then 



-^- + -21 = -*- (p + q) 
200 200 200 

GL = M + /(< ?-P> + ^P = s <P + g) 
4 x 200 200 400 



/ = 



\ s(p + q) _ JW _ M ] ^ <7 - P 
L 400 200 J ' 800 



2sp + 2sq - 4sp ~ 800 M 

q - P 
2s(q-p) - 800JW 



= 2s - 



q - p 
800 M 



If AL = LE, then 

M = H 



q-p 

h, + h 2 



and 



/ = 2s - 800 Iff - fc i+M (12.15) 



(2) s 4 I 



In Fig. 12.8, let the depth of the curve below the centre of 
the chord T, T 2 = M 

u = ml = fo ~ p) s * 

4 800/ 

/. / = s2 ( q ~ P^ 
800 M 



VERTICAL AND TRANSITION CURVES 



'<q-p) 



615 
(12.16) 



800 



(H-hl±hlj 



I = 



If S = /, 

q - p 



(2) S« 



(12.17) 




Fig. 12.8 

12.43 Sight distance related to the length of the beam of a vehicle's 
headlamp 

(1) s > I 

In Fig. 12.9, the height of the beam = h is at A, 

the beam hits the road at T 2 , 

the angle of the beam is $° above the horizontal 

axis of the vehicle. 



S>1 

HB, s, Is assumed equal to s 




Fig. 12.9 Sight distance related to 
the beam of a headlamp 



In triangle BT 2 0, 



i.e. 



T 2 Q = sO 

al z - h = s6 



616 SURVEYING PROBLEMS AND SOLUTIONS 

But a = q ~ P 

200/ 

(< ?-P> /2 = sd + h 
200/ 



/ = 200(s^) 



Q 
Practice suggests that 6 = 1° 

h = 2-5 ft 

Then / = 200(0-0175 s+ 2' 5) 3'5s + 500 

9 - P <7 - P 

(2) s«Z 

As before, as 2 - h = sd 

200/ ~ s6 + h 

I = s2 (g ~ P) 
200(s<9 + a; 
If 0=1° and h = 2-5 ft 



(12.19) 



200(s<9 
1° 


+ h) 
and 


s 2 (g- 


-P) 



(12.20) 



/ = J ^ ~ ^ (12 21) 

3-5 s + 500 K ' 

Example 12.2. The sag vertical curve between gradients of 3 in 100 

downhill and 2 in 100 uphill is to be designed on the basis that the 

headlamp sight distance of a car travelling along the curve equals the 

minimum safe stopping distance at the maximum permitted car speed. 

The headlamps are 2'5ft above the road surface and their beams tilt 

upwards at an angle of 1° above the longitudinal axis of the car. The 

minimum safe stopping distance is 500 ft. 

Calculate the length of the curve, given that it is greater than the 

(L.U.) 



sight distance. 






p = -3% 
q = +2% 
s = 500 ft 




By Eq. (12.21), 










/ = 


s 2 (q- 


P) 


500 2 x (2+3) 


500 x 5 




3*5 s + 


500 


500 x 3*5 + 500 


3-5 + 1 


= 


2500 
4-5 


555- 5 ft 





12.5 Setting-out Data 

Gradients are generally obtained by levelling at chainage points, 
e.g. A,B,C and D, Fig. 12.10. 



VERTICAL AND TRANSITION CURVES 



617 



+p<y 


~~^~^^qVo 


^( 




r, 


'/2/ 


X 


d-x 


'••^^7' 


2 




d 






1/2/ 











Si. A St.S St.C 

Levels taken at stations A,B,C and 

Fig. 12. 10 Setting out data 

Chainage and levels of I, T, and T 2 . 

Level of / = level of B + £— 

100 

= level of c + \Za 

Solving the equation gives the value of x. 
Chainage of / = chainage of B + x 

Chainage of T, = chainage of / - - / 
Chainage of T 2 = chainage of / + ^ / 



Level of T, = level of / - 
Level of T = level of / - 



lp 
200 

200 



Levels on the curve (Fig. 12.11) 



st.o 



(12.22) 

(12.23) 
(12.24) 

(12.25) 

(12.26) 

(12.27) 









A/ 










. 


r~ter~~— — 




jp<^ 


r^~ 


i 








r v^i 






!c 









-<7% 



(a) Summit 
BC = tox-ax 2 



Fig. 12.11 Levels on the curve 



(b) Valley 
BiC y =-bx*ax 2 




618 SURVEYING PROBLEMS AND SOLUTIONS 

Levels on the tangent at A = level of T, + bx (12.28) 
where b = ±p/100. 

Difference in level between tangent and curve = ±ax 2 (12.29) 
Levels on the curve at B = level of tangent + difference 

in level between curve and tangent 
= T, ± AC + AB (12.30) 

Check on computation 

(a) Tangent level is obtained by successive addition of the dif- 
ference in level per station. 

(b) Successive curve levels should check back to the level obtain- 
ed by the spot level derived from y = + ax 2 ± bx 

(c) The final value of the check on T 2 will prove that the tangent 
levels have been correctly computed, though the curve levels may not 
necessarily be correct. 

In order to define the shape of the curve, the values of a and b 
in the formula must be in some way obtained, 
p and q will always be known. 

b = p/100 
Either l,s or a must be given, and if the value of s is required, 
the height of the vehicle (h) must be known. For general purposes 
this is taken as 3*75 ft, and Ministry of Transport Memoranda on recom- 
mended visibility distances are periodically published. 

Example 12.3. As part of a dual highway reconstruction scheme, a 
line of levels were taken at given points on the existing surface. 

Reduced level Chainage 

A 104-63 20 + 75 

B 109-13 22 + 25 

C 107-29 25 + 50 

D 103-79 27 + 25 

If the curve, based on a simple parabola, is designed to give a 
rate of change of gradient of 0-6% per 100 ft, calculate: 

(a) the length of the curve /, 

(b) the chainage and level of the intersection point, 

(c) the chainage and level of the tangent points, 

(d) the level of the first three chainage points on the curve (i.e. 
stations 100 ft apart based on through chainage), 

(e) the length of the line of sight s to a similar vehicle of a 
driver 3 ft 9 in. above the road surface. 

(N.B. s < /) 



VERTICAL AND TRANSITION CURVES 



619 




104-63 
2075 



109-13 
2225 



107-29 103-79 levels 

2550 2725 chainage 



Fig. 12. 12 



(a) Gradient AB = (109-13 - 104-63) in (2225 - 2075) 

= 4-50 ft in 150 ft 
i.e. +3% 
CD = (107-29 - 103-79) in (1725 - 2550) 
= 3-50 ft in 175 ft 
i.e. -2% 

By Eq. (12.6), Length of curve = 100(3 + 2) = 333.33 ft 

0-6 



(b) By Eqs. (12.22/23), 

Level of / = 109-13 + — 

100 



= 107-29 + 2 ( 325 ~*) 
100 



Solving for x, 



x = 93-20 ft 
325 - x = 231 -80 ft 



Level of / = 109-13 + 3 x 93 ' 2 = 111-93 



also 



100 

= 107-29 + 2_*-?31± 
100 



= 111-93 {check) 



Chainage of / = Chainage of B + x 
= (22 + 25) + 93-20 
= 23 + 18-20 
(c)By Eqs. (12.24/25), 

Chainage of T, = " 2318-20 - 1/2 

= 2318-20 - 416-67 = 1901-53 ft 
Chainage of T 2 = 2318-20 + 416-67 = 2734-87 ft 



620 



SURVEYING PROBLEMS AND SOLUTIONS 



By Eqs. (12.26/27), 



Level of T, = 111-93 - 3 x 833 ' 33 = 99-43 ft 

1 200 

Level of T 2 = 111-93 - 2 x * 33 ' 33 

2 200 



103-60 ft 



(d) Setting-out data 

ax 2 
Point Chainage Length (x) Tangent Level u Formation 

(0-3xlO-V) Level 
T, 1901-5 99-43 99-43 

+ 2-955 
2000-0 98-5 102-385 -0-291 102-09 

+ 3-000 
2100-0 198-5 105-385 -1-181 104-20 

+ 3-000 
2200-0 298-5 108-385 -2-675 105-71 

(e) Line of sight (s) 



,._o»/>i 



In Fig. 12.13, 



but 



Fig. 


12 


13 


h , = adf 




d - IEL 

1 - Va 


h 2 = a d z 




d 2 = M 
2 V a 


s = d, + d 2 




V7 + 'V a 



h, = h z a = 0-3 x 10" 

s = 2 IK = 200 A /S? 
'V a ^0-3 

= 707 ft 



Example 12.4. A 6% downgrade on a proposed road is followed by a 
1% upgrade. The chainage and reduced level of the intersection point 
of the grades is 2010 ft and 58-62 ft respectively. A vertical parabolic 
curve, not less than 250 ft long, is to be designed to connect the two 
grades. Its actual length is to be determined by the fact that at chain- 
age 2180 ft, the reduced level on the curve is to be 61-61 ft to provide 
adequate headroom under the bridge at that point. 



VERTICAL AND TRANSITION CURVES 



621 



Calculate the required length of the curve and also the chainage 
and reduced level of its lowest point. (R.I.C.S.) 




In Fig. 12.14, 



Thus, 



Fig. 12.14 



let x -I = 170ft 
2 



level at / = 58*62 

tangent level at Y = 58*62 - (170 x 0*06) 

= 58-62 - 10*20 = 48-42 ft 

curve level at X = 61*61 ft 

.-. XY = 13* 19 ft 

ax 2 = 13*19 

13* 19 

i.e. a = 



Also, 



Thus 



al 2 = -EL + JL = JL(p +q) 
200 200 200^ Hf 



a = P + = -J— 
200/ 200/ 

13-19 



200/ 



7x'< 



Thus 

i.e. 

Solving for x, 



x - 



7x' 



200 x 13*19 
170 



7x z 
2638 



5276 
7x2 - 5276 x = 170 x 5276 

+ 5276 ± V15276 2 - 4 x 7 x 170 x 5276 j 
14 

= 5276 ± 1649*9 = 494 . 7 or 259 . Q ft 
14 



622 SURVEYING PROBLEMS AND SOLUTIONS 



i.e. 


/ 
2 


324-7 


or 89-0 ft 






/ = 


650 ft 


(length must not b« 








250 ft) 




To find the 


minimum height 


on the curve, 








H 


= aX z - bX 

_ (p + q) X Z 
200/ 


pX 
100 






dH 
dX 


_ 2(p + q) X 
200/ 


p 

100 




.-. 


X 


pi 





_iL = 



p + q 

557-1 ft 



6 x 650 



Chainage of minimum height = 2010 + 557-1 - —— 

= 2242-1 ft 



Level at minimum height: 

Level at 7, = 58- 62 + 0-06 x (325) = 78- 12 ft 
Level of tangent = 78*12 - 0-06 x (557- 1) = 44*70 



i2 



aX 2 = L* 55 !'i = 16-71 ft 
200 x 650 

Level on curve = 44-70 + 16-71 
= 61-41 ft 

Example 12.5. On the application of the cubic parabola for valley 
curves. 

A valley curve of length 400 ft is to be introduced into a road to 
link a descending gradient of 1 in 30 and an ascending gradient of 1 in 
25. It is composed of two cubic parabolas, symmetrical about the bi- 
sector of the angle of intersection of the two straights produced. 

The chainage of the P.I. of the straights is 265 + 87 ft, and its re- 
duced level 115-36 ft. 

Calculate: 

(i) The reduced levels of the beginning, the mid-point and the end 
of the curve. 

(ii) The chainage and reduced level of the lowest point of the 
curve. 

(iii) The reduced level at chainage 267 + 00 ft. 



VERTICAL AND TRANSITION CURVES 



623 



(The formula for the cubic parabola relating to the curve of total length 

L and terminal radius R is 3 \ 

y = _*_] (N.U.) 

6RL/ 




263+87 



Fig. 12. 15 

From the gradients, Fig. 12.15, 0, = cot _, 30 = 1°54'33" 

0, = cot _1 25 = 2°15'27" 



<f> = |[180- (0, + M = 87°55'00" 



6KL 

i£l 
6RL 



From the equation, Y 

dy 
dx 

and if x = L = 400/2, then K = 100 tan<£ 

= 100 x 27-49 = 2749 ft 



cot<£ 



At I, y = 



6R 



40000 
6 x 2749 



- 2-43 



(i) As the gradients are small, the values of y are assumed vertical. 
.*. Level of curve at the mid point = 115*36 + 2* 43 = 117*79 ft 

Level at T, = 115*36 + ^ = 115-36 + 6*67 = 122*03 ft 
Level at T 2 = 115*36 + ^ = 115*36 + 8-00 - 123*36 ft 



(ii) Level at the lowest point P 

(relative to T,) 



— x x 

30 + 6RL 



dL = -_L + _*_ 
30 2/?L 



dx 



2RL 
30 



= (for min value) 

2 x 2749 x 200 



30 



x = 191 ft from I, 



624 SURVEYING PROBLEMS AND SOLUTIONS 

The chainage of the lowest point P 

= chainage of T, + x 

= (263 + 87) + (1 + 91) = 265 + 78 ft 

Level of P = - ^ + ^ - + 122-03 

30 6 x 2749 x 200 

= -6-37 + 2-11 + 122-03 = 117-77 ft 
(iii) At chainage 267 +00, i.e. 87ft from T 2 , 



87 



y = 2i = 0-199 ft, i.e. 0-20 ft 

6 x 2749 x 200 

.'. Level at 267 + 00 = 123-36 - -^ + 0-20 

25 

= 120-08 ft 



Exercises 12(a) 

1. An uphill gradient of 1 in 100 meets a downhill gradient of 0*45 
in 100 at a point where the chainage is 61+00 and the reduced level 
is 126 ft. If the rate of change of gradient is to be 0*18 % per 100 ft, 
prepare a table for setting out a vertical curve at intervals of 100 ft. 

(I.C.E. Ans. 121-97, 122-88, 123-61, 124-16, 124-53, 
124-72, 124-73, 124-56, 124-19) 

2. (a) A rising gradient of 1 vertically to 200 horizontally, is to be 
joined by a rising gradient of 1 in 400 by a 400 ft long parabolic curve. 
If the two gradients meet at a level of 365-00 ft A.O.D., tabulate the 
levels on the curve at 50 ft intervals. 

(b) Recalculate on the basis that the first gradient is falling and 
the second likewise falling in the same direction. 

(L.U. Ans. 364-00, 364-24, 364-47, 364-68, 364-87, 

365-05, 365-22, 365-37, 365-50; 

366-00, 365-76, 365-53, 365-32, 365-13, 

364-95, 364-78, 364-63, 364-50) 

3. A rising gradient, g,, is followed by another rising gradient g 2 
(g 2 less than g,). These gradients are connected by a vertical curve 
having a constant rate of change of gradient. Show that at any point 
on the curve, the height y above the first tangent point A is given by 

y = g\X - ^ gl ~ ? f z where x is the horizontal distance of the point 

from A y and L is the horizontal distance between the tangent points. 
Draw up a table of heights above A for 100 ft pegs from A, when 
g, = +5%, g 2 = +2% and L = 1000 ft. At what horizontal distance from 
A is the gradient +3%? 



VERTICAL AND TRANSITION CURVES 625 

(I.C.E. Ans. 4-85, 9-40, 13-65, 17-60, 21-25, 24-60, 27-65, 

30-40, 32-85, 35-00, 667 ft) 

4. A rising gradient of 1 in 100 meets a falling gradient of 1 in 150 
at a level of 210*00. Allowing for headroom and working thickness, 
the vertical parabolic curve joining the two straights is to be at a level 
of 208-00 at its midpoint. 

Determine the length of the curve and the levels at 100 ft intervals 
from the first tangent point. 

(L.U. Ans. 960 ft; 200-40, 206-11, 206-85, 207-42, 207-81, 
208-00, 208-03, 208-07, 207-94, 207-63, 207-15, 206-80) 

5. On a straight portion of a new road, an upward gradient of 1 in 100 
was connected to a gradient of 1 in 150 by a vertical parabolic summit 
curve of length 500 ft. A point P, at chainage 59 100 ft on the first 
gradient, was found to have a reduced level of 45-12 ft, and at point Q, 
at chainage 60000 ft on the second gradient, of 44*95 ft. 

(a) Find the chainages and reduced levels of the tangent points 
to the curve. 

(b) Tabulate the reduced levels of the points on the curve at inter- 
vals of 100 ft from P at its highest point. 

Find the minimum sighting distance to the road surface for each of 
the following cases: 

(c) the driver of a car whose eye is 4 ft above the surface of the 
road. 

(d) the driver of a lorry for whom the similar distance is 6 ft. 
(Take the sighting distance as the length of the tangent from the 

driver's eye to the road surface.) 

(L.U. Ans. (a) 59 200, 46-12, 59 700, 46*94 

(b) 46-12, 46-95, 47-45, 47-62 (highest point) 
47-45, 46-94. 

(c) 605 ft (d) 845 ft 

6. A rising gradient of 1 in 500 meets another rising gradient of 1 in 
400 at a level of 264-40 ft, and a second gradient 600 ft long then meets 
a falling gradient of 1 in 600. The gradients are to be joined by two 
transition curves, each 400 ft long. 

Calculate the levels on the curves at 100 ft intervals. 
(L.U. Ans. 264-00, 264-21, 264*42, 264-65, 264-90, 265-15, 
265-40, 265*61, 265-70, 265*69, 265-58) 

7. A falling gradient of 4% meets a rising gradient of 5% at chainage 
2450-0 ft and level 216-42 ft. 

At chainage 2350, the underside of a bridge has a level of 235-54 
ft. The two gradients are to be joined by a vertical parabolic curve 
giving 14 ft clearance under the bridge. List the levels at 50 ft inter- 
vals along the curve. 



626 SURVEYING PROBLEMS AND SOLUTIONS 

(L.U. Ans. 224-42, 222-70, 221-54, 220-95, 220-92, 221-45, 

222-54, 224-20, 226-42) 

8. The surface of a length of a propsed road of a rising gradient of 
2% is followed by a falling gradient of 4% with the two gradients joined 
by a vertical parabolic summit curve 400 ft long. The two gradients 
produced meet at a reduced level of 95*00 ft O.D. 

Compute the reduced level of the curve at the ends, at 100 ft inter- 
vals, and at the highest point. 

What is the minimum distance at which a driver whose eye is 3 ft 

9 in. above the road surface would be unable to see an obstruction 4 
inches high ? 

(I.C.E. Ans. 91-00, 92*25, 92-00, 90*25, 87*00 ft A.O.D. 

highest point 92-33 ft AOD ; 
sight distance 290 ft) 

9. An existing length of road consists of a rising gradient of 1 in 20, 
followed by a vertical parabolic summit curve 300 ft long, and then a 
falling gradient of 1 in 40. The curve joins both gradients tangent- 
ially and the reduced level of the highest point on the curve is 
173*07 ft above datum. 

Visibility is to be improved over the stretch of road by replacing 
this curve with another parabolic curve 600 ft long. 

Find the depth of excavation required at the midpoint of the curve. 
Tabulate the reduced levels of points at 100 ft intervals on the new 
curve. 

What will be the minimum visibility on the new curve for a driver 
whose eyes are 4*0 ft above the road surface? 

(I.C.E. Ans. 2*81 ft; 160-57, 164-95, 168-08, 169-95, 170-61, 

169-99, 168*07 ft A.O.D. ; 
minimum visibility 253 ft) 

10 A vertical curve 400 ft long of the parabolic type is to join a fall- 
ing gradient of 1 in 200 to a rising gradient of 1 in 300. If the level of 
the intersection of the two gradients is 101*20 ft, give the levels at 50 
ft intervals along the curve. 

If the headlamp of a car was 1*25 ft above the road surface, at what 
distance will the beam strike the road surface when the car is at the 
start of the curve ? Assume that the beam is horizontal when the car is 
on a level surface. 

(L.U. Ans. 102-20, 101-98, 101-80, 101-68, 101-62, 101-60, 

101-64, 101*72, 101*86; 347ft) 

11. A vertical parabolic curve 500 ft long connects an upward gradient 
of 1 in 100 to a downward gradient of 1 in 50. If the tangent point T, 
between the first gradient and the curve is taken as datum, calculate 
the levels of points at intervals of 100 ft along the curve, until it meets 



VERTICAL AND TRANSITION CURVES 627 

the second gradient at T z . 

Calculate also the level of the summit giving the horizontal dis- 
tance of this point from T,. 

If an object 3 in. high is lying on the road between T, and T 2 at 
10 ft from T 2 and a car is approaching from the direction of T,, calcu- 
late the position of the car when the driver first sees the object, if his 
eyes are 4 ft above the road surface. 

(L.U. Ans. 0-70, 0-80, 0*30, -0*80, -2-50, 0'84, 166-67 ft 

33-6 ft from T,) 

12. A parabolic vertical curve of length L is formed at a summit 
between an uphill gradient of a% and a downhill gradient of b%. 
As part of a road improvement, the uphill gradient is reduced to c % 
and the downhill gradient increased to d%, but as much as possible of 
the original curve is retained. 

Show that the length of the new vertical curve is 

L x (c + d) (L.U.) 

(a + b) 

13. The algebraic difference in the gradient of a sag vertical curve 
L ft long is a ft/ ft. The headlamps of a car travelling along this curve 
are 2*5 ft above the road surface and their beams tilt upwards at an 
angle of 1° above the longitudinal axis of the car. 

Show that if s, the sight distance in feet, is less than L, then, 

L = ^ (L.U.) 

5 + 0-035 s 

12.6 Transition Curves 

12.61 Superelevation (0) 

A vehicle of weight W (lbf) or (kgf) on a curve of radius r (ft) or 
(m), is travelling at a velocity v (ft/s) or (m/s) or V (mile/h). Fig. 
12.16 shows the centrifugal force Wv 2 /gr which must be resisted by 
either (a) the rails in the case of a railway train, or (b) adhesion be- 
tween the road and the vehicle's tyres, unless superelevation is applied, 
when the forces along the plane are equalised. 

Then Wv 2 cosfl = Wsind (12.31) 

gr 



i.e. 

If 6 is small, 



tan0 = Z_ (12.32) 

gr 



rad = Z_ (known as the centripetal 
8 r ratio) 



(12.33) 



628 



SURVEYING PROBLEMS AND SOLUTIONS 




cos 6 
an 

W sin 6 
Fig. 12.16 
12.62 Cant (c) 

If d is the width of the track, then the cant c is given as 

c = dsind (12.34) 



dv 2 
gr 



COS0 



~ ^l! (if e is small) 
gr 



(12.35) 
(12.36) 



N.B. c oc v 2 
oc 1/r 

On railways c is usually limited to 6 in. with a 4 ft 8y2in. gauge. 
On roads tantf is usually limited to O'l. 



12.63 Minimum curvature for standard velocity 

Without superelevation on roads, side slip will occur if the side 
thrust is greater than the adhesion, i.e. if Wv z /gr > /j,W, where jjl ~ 
the coefficient of adhesion, usually taken as 0'25. 



Thus the limiting radius r = Z_ 
If the velocity V is given in mile/h, 



(12.37) 



r = 



ft 



15 



M 



(12.38) 



VERTICAL AND TRANSITION CURVES 629 

12.64 Length of transition 

Various criteria are suggested: 

(1) An arbitrary length of say 200 ft. 

(2) The total length of the curve divided into 3 equal parts, 1/3 
each transition; 1/3 circular. 

(3) An arbitrary gradient of 1 in. in sft, e.g. lin. in 25 ft -the 
steepest gradient recommended for railways. 

(4) At a limited rate of change of radial acceleration (W.H. Short, 
1908, fixed 1 ft/s 3 as a suitable value for passenger comfort.) 

(5) At an arbitrary time rate i.e. 1 to 2 in per second. 
N.B. (4) is the most widely adopted. 

12.65 Radial acceleration 

The radial acceleration increases from zero at the start of the 
transition to v 2 /r at the join with the circular curve. 

The time taken t = — , where / = the length of 

v 

transition. 

.". the rate of gain of radial acceleration 

r ' v 

,3 



- 1- ft/s 3 
rl 


(12.39) 


Thus the length of the curve I is given as 




l - 21. 

ar 


(12.40) 


If a is limited to 1 ft/s 3 




then / = ll (ft) 

r 


(12.41) 


3-155 V 3 (ft) 


(12.42) 



On sharp radius curves, the value of a would be too great, so the 
superelevation is limited and the speed must be reduced. 

From Eq. (12.32), ■ z 

tan = _ 
gr 

then v = y/(g.r tan<9) (12.43) 

v 3 

but 1 = 1- 

ar 



630 SURVEYING PROBLEMS AND SOLUTIONS 

/ = (stan<9) 3/2 ^ (12.44) 

a 

i.e. when c is limited, / oc \Jr. 

For railways with a maximum superelevation of 6 in., 

sinfl = Ac 6 '™' = 0-1008 
4 ft SVi in. 

6 = 5° 47' 

tan# = 0-1013. 

Thus the maximum speed should be: 

v = VC 32 ^ x 0-1013 r) 

= 1-806 Vr ft/s (12.45) 

^ 2y/r ft/s (12.46) 



(12.47) 



(12.48) 

12.7 The Ideal Transition Curve 

If the centrifugal force F = Wv 2 /gr is to increase at a constant 

rate, it must vary with time and therefore, if the speed is constant, 

with distance. , 

i.e. r oc / oc 

gr 

I oc — i.e. rl = constant '= RL 
r 

where R = the radius of the circular curves 

L = the total length of the transition. 

In Fig. 12.17, 

81 = rS<£ 

d<f> = -dl 
r 

but rl = RL = constant k 

dcf> = -j-dl 



If a 


= If 


t/s 3 , 






/ 


r 
= Sy/r 


(2 V / r) 3 
r 


If / 


and 


r are 


given 


in 


Guntei 


r chains, 














66/ 


= 8V(66r) 












/ 


* V 


chains 



Integrating, , / 2 



k 
2k 



VERTICAL AND TRANSITION CURVES 



631 




Fig. 12. 17 The ideal transition curve 

but </> = when / = 

c = 



= 



/ = 



r 



2RL 

WV0 



where M = y/2RL 



(12.49) 



(12.50) 



This is the intrinsic equation of the clothoid, to which the cubic 
parabola and lemniscate are approximations often adopted when the 
deviation angle is small, Fig. 12.18. 





y 




(a) Clothoid 

(b) Lemniscate 

(c) Cubic parabola 




c 


"o 1 




X 



Fig. 12. 18 



(a) Clothoid I = My/</> 

(b) Lemniscate c 2 = a 2 sin 20 

x 3 



(c) Cubic parabola y = 



6RX 



(y = — — cubic spiral ) 
V 6RL I 



632 



SURVEYING PROBLEMS AND SOLUTIONS 



12.8 The Clothoid 



/ = V(2/?L)V0 
- l z /2RL 

where R = the minimum radius (i.e. the radius of the circular curve). 
As the variable angle <f> cannot be measured from one position, it 
is difficult to set out in this form. 




Fig. 12. 19 Clothoid 



12.81 To find Cartesian co-ordinates 



dx 
dl 



= cos 



2! 4! 



= 1 - 



' + *' 



Integrating, 



[ l A 

Y ~ 5x2!(2 

■i 



2! (2RL) 2 4! (2KL) 4 

_ / 8 

(2RL) 2 + 9x4!(2RL) 4 ' 



1 - ^ +_£_ 
5x2! 9x4! 



For <£max I = L 

then x ~ / 



I 



40K 2 J 



Similarly, 



=¥. = sin <f> = $ - -$- + *£— 



dl 



3! 5! 



r 



/ 6 / ,0 

+ 



Integrating, 



y = / 



r 



2RL 3!(2KL) 3 5!(2KL) 5 

z 6 / ,0 



3(2RL) 7x3!(2RL) 3 llx5\(2RLf 
<f> 3 



- iTi _ _*!_ + _5^! 1 

1.3 7x3! 11x5! J 



(12.51) 
(12.52) 

(12.53) 
(12.54) 

(12.55) 

(12.56) 
(12.57) 

(12.58) 
(12.59) 



VERTICAL AND TRANSITION CURVES 



633 



For <£ ma3t , 



6R [ 56R 2 



12.82 The tangential angle a 



Tana = X. = ± + _£! + 
x 3 105 



but 

By substitution, 



a = tana - -|tan 3 a + Itan 5 a 
•3 5 



a = ±- Ml 
3 2835 



(12.60) 

(12.61) 
(12.62) 
(12.63) 



i.e. a = & - k (a rapidly decreasing (12.64) 



Thus, if <p is small, 



quantity) 



a = 2- 



(12.65) 



Jenkins* shows that if <f> < 6°, no correction is required; and if 
< 20° no correction > 20" is required. 

12.83 Amount of shift (s) 

The shift is the displacement of the circular curve from the 
tangent, i.e. DF, Fig. 12.20. 




T F G H 

Fig. 12. 20 Amount of shift 



ByEq.(12.59), PH = BF = y = l\^-— ^ + ...1 

L 3 7x3! J 

DF = BF - BD = y - fl(l- cos<£) 

J 2 T 

Expanding cos^ and putting max = ^ = ^ 



DF = 

24R 

* R. B.M.Jenkins, Curve Surveying (Macmillan). 






(12.66) 
(12.67) 

(12.68) 



634 



SURVEYING PROBLEMS AND SOLUTIONS 



12.9 The Bernouilli Lemniscate 

Polar equation c 2 = a 2 sin 2a 

Identical to clothoid for deviation angles up to 60° (radius de- 
creases up to 135°), Fig. 12.21, 

c is maximum = a when a = 45° 








*\ \ 


*1 
>l 

L 
31 






B 


\r=r« 








J^^y 








^g£s£^ 


0-V 
e A. 


\135 4 



r, F N 

Fig. 12.21 Principles of the Bernouilli lemniscate 



Referring to Fig. 12.22, 



2c 2 gives - 



c 2 = a 2 sin 2a 

2c — = 2a 2 cos 2a 
da 

1 dc 2a 2 cos 2a 



(12.69) 



cot 2a 



c da 2a 2 sin 2a 
If $ = angle T X PQ and P, and P 2 are 2 neighbouring points, 

cot0 = ^ (P 2 MP, =90°) 
P 2 lw 

5c 



c8a 



i.e. 



Iff£ = cot0 
c da 



:. = 2a 

<f> = + a = 3a 



as shown in the clothoid and the cubic parabola. 



(12.70) 



p,p 2 - a - 



cSa 



VERTICAL AND TRANSITION CURVES 

P 2 M c8a 



635 



sin 6 sin 



6* 



sin 2a 






. dl c c 


c 


\ \Sr^ 


da sin 2a c 2 /a 2 


^jjz^r^ 


Now d/ = rd<£ = 3rda 




See c^- — ^^\^r 


da c 




^s^ >^* 


Fig. 12.22 


Thus 


a 2 = 


= 3rc (12.71) 


i.e., from Eq. (12.69) 


c 2 = 


= 3Rc sin 2a if the lemniscate 

approximates to the 
circle radius R 




c = 


= 3/? sin 2a (12.72) 



In Fig. 12.21 



OF = OB + BF 

= r cos <f> + c sin a 

= r cos0 + 3r sin a sin 2a 

= r[cos3a + 3 sin a sin 2a] 



(12.73) 



T,F = T,N - FN 

= c cos a - r sin 

= 3r sin 2a cos a - r sin 3a 

= r[3 sin 2a cos a - sin 3a] 



(12.74) 



12.91 Setting out using the lemniscate 

Using Eq. (12.72), 

c = 3R sin 2a 

Shift DF = OF - OD 

= R [cos 3a + 3 sin a sin 2a] - R 

= /?[cos3a + 3 sin a sin 2a - 1] (12.75) 

Equal values of a enable values of c to be computed, or equal chords 
c, 2c, 3c, etc. 

sin 2a = JL (12.76) 

3R 



where a = $ 



636 SURVEYING PROBLEMS AND SOLUTIONS 

If a is small, a" = 206265c 

6R 



(12.77) 



<W - ^f^ (12.78) 

(The same value as with the cubic spiral) 
For offset values, 

y = c sin a (12.79) 

x = c cos a (12.80) 

If the chord lengths are required between adjacent points on the 
curve, the length of the chord 

c = y/{x 2 +y 2 ) (12.81) 

12.10 The cubic parabola 

This is probably the most widely used in practice because of its 
simplicity. It is almost identical with the clothoid and lemniscate for 
deviation angles up to 12°. The radius of curvature reaches a mini- 
mum for deviation angles of 24° 06' and then increases. It is therefore 
not acceptable beyond this point. 

.3 



Let y = A_ 

k 




(12.82) 


dy _ 3x 2 
dx k 




(12.83) 


d y _ 6x 






dx 2 k 




(12.84) 


By the calculus the curvature (p) is given as 






1 d 2 y/dx 2 

p = — = il 


3/2 


(12.85) 



( + \dx) 



If <f> is small, dy/dx is small and (dy/dx) is neglected. 

jl_ _ d y _ 6x 

7 dx 2 ~ ~k~ 



(12.86) 



k = 6rx = 6RX at the end of (12.87) 

Equation of the cubic parabola e curve 

y = -^- (12.88) 

6RX 

If the deviation angle <f> is small, x ~ /, X o- L. 

Equation of the cubic spiral 

y - <£ (1289) 



VERTICAL AND TRANSITION CURVES 637 

N.B. This is the first term in the clothoid series. 



cf> ~ tan<£ 


_ dy _ 3x 2 
dx 6RX 


x 2 

2RX 


(12.90) 


a ^ tana 


_ y _ x 2 
x 6RX 




(12.91) 


.\ a 


3^ 




(12.92) 



as in the first term of the clothoid series. 
In Fig. 12.21, 

PB = /? sin <£ ^ /?<£ = X/2 (12.93) 

i.e. the shift bisects the length. 

DB = R(l-cos<f>) (12.94) 

Shift DF = y-/?(l-cos<£) (12.95) 



X a on _:_, ' 



6RX 



- 2R sin* - 



X 2 


2R<f> 2 


6/? 


4 


x 2 

6/? 


x 2 

8/? 


X 2 
24R 


L 2 
24/? 



the first term in the Clothoid series 
As E is on transition and TF = FH = kx, 



EF 



%' 



X 3 



X 2 



(12.96) 



6RX 48/?X 

\DF (12.97) 



48/? 
i.e. the transition bisects the shift. 

12.11 The Insertion of Transition Curves 

The insertion of transition curves into the existing alignment of 
straights is done by one of the following alternatives. 

(1) The radius of the existing circular curve is reduced by the 
amount of 'shift', Fig. 12.23. The centre is retained. 

/?, - R 2 = shift (5) = -L_ 
2 24/? 

T'T if 

1 1 * 1 a> 21-, 



638 



SURVEYING PROBLEMS AND SOLUTIONS 




Fig. 12.23 

(2) The radius and centre are retained and the tangents are 
moved outwards to allow transition, Fig. 12.24. 
(N.B. Part of the original curve is retained.) 

/,/ 2 = 0I 2 - 0/, 

= R + S R 

cos — COS — 

2 2 



cos 




Fig. 12.24 



VERTICAL AND TRANSITION CURVES 



639 



(3) The radius of the curve is retained, but the centre is moved 
away from the intersection point, Fig. 12.25. 

o,o 2 = shift 



cos 



u 




Fig. 12.25 

(4) Tangent, radius and part of the existing curve are retained, but 
a compound circular curve is introduced to allow shift, Fig. 12.26. 




Fig. 12.26 
(5) A combination of any of these forms. 



640 



SURVEYING PROBLEMS AND SOLUTIONS 



12.12 Setting-out Processes 

Location of tangent points (Fig. 12.27) 




TyP, Transition curve 

PiP a Circular curve 

%Ti — Transition curve 



Fig. 12.27 Setting-out processes 



By Eq. (12.96), 



FD 



shift (s) = ±L 



24R 

FI = (R + s) tan iA 
FT ^ H 
Tangent length I,/ = T 2 I = ^L + (/? + s) tan 5A (12.98) 

Setting out (Fig. 12.27) 

(1) Produce straights (if possible) to meet at /. Measure A. 

(2) Measure tangent lengths /T, = IT 2 to locate T, and T 2 . 

(3) Set out transition from T, . Two methods are possible, (a) by 
offsets from the tangent, (b) by tangential angles. For either method, 
accuracy is reduced when 

/ > 0-4R i.e. <f> > 12° 

Method (b) is more accurate, even assuming chord = arc. 



VERTICAL AND TRANSITION CURVES 



641 



(4) Offsets from the tangent (Fig. 12.28) 



From Eq. (12.88), 

~3 



y, = 



6RL 

IX X 2 = «X < 9 X g = ox ^ 

then y 2 = 2 3 y, = 8y, 

y 3 = 3 3 y, = 27y, 




Fig. 12.28 Offsets from the tangent 



(5) Tangential angles a (Fig. 12.29) 




Fig. 12.29 Deflection angles from the tangent 
From Eq. (12.89) 



tana 



If a is small, then 
if x ^ c 



a = 



6RL 

206 265 x 2 
6RL 

206 265 c 2 
6RL 



(12.99) 
(12.100) 



For a" c^ L 

max — 



a' ^ 573 c 2 /RL 



OL_ = 



206 265 c 
6R 



a' = 573 c/R 



(12.101) 
(12.102) 



642 SURVEYING PROBLEMS AND SOLUTIONS 

0m«* = 3a m = 206 265c/2K (12.103) 

<f>' m = 1719 c/R (12.104) 

(6) Check on P, (Join of transition to circular curve) 

N,P, = N 2 P 2 = ^ (12.105) 

(7) Move theodolite to P,. 

Set out circular curve by offsets or deflection angles from the tan- 
gent QPZ. 
N.B. Angle T,P,Q, = 6 = 2a = §0 m (12.106) 

Checfc ZP,P 2 = /8 = | A - <£ m (12.107) 

P,P 2 = 2R sinjS (12.108) 

Check N 2 P 2 = N,P, = ^ Eq. (12.105) 

(8) Move theodolite to T 2 and set out the transition backwards towards 
P 2 as in (3). 

N.B. The use of metric units does not make any difference to the solu- 
tion, providing these are compatible, i.e. 

v - m/s 

L and R - m 

a - ft/s 3 converted to m/s 3 

(e.g. lft/s 3 = 0-305 m/s 3 ) 

Example 12.6. A circular curve of 2000 ft radius deflects through an 
angle of 40° 30'. This curve is to be replaced by one of smaller radius 
so as to admit transitions 350 ft long at each end. The deviation of the 
new curve from the old at their midpoint is 1*5 ft towards the intersect 
tion point. 

Determine the amended radius assuming the shift can be calculated 
with sufficient accuracy on the old radius. Calculate the length of the 
track to be lifted and the new track to be laid. 

(L.U.) 

By (12.94) Shift ,, . JL . ^|5l_ .. 2 . 55 ft . 

Tangent length of circular curve (7J /) = 2000 tan40°30^- = 737*84 ft. 

In triangle 0,/T, ftl . -JSSJL-, . 2131 . 8ft 

X,/ = 2131*8 - 2000 = 131 -8 ft 
new value XJ = 131*8 - 1*5 = 130*3 ft 



VERTICAL AND TRANSITION CURVES 



643 



Not to scale 



In triangle OJF 




A - 40* 30' 



= cos 20° 15' = 0*93 819 



i.e. 



By (12.103) 



By (12.107) 



R + 130-3 

K + 2-55 = (R + 130-3)0-93819 

•• R = Trim 5 ■ i2*Lift 

206265x350 1fi ,, Q „ 
*- = 3a = 2 x 1936-5 = 18639 

= 5° 10' 39" 

= iA - <£ max 

= 20° 15' - 5° 10' 39" = 15° 04' 21" 



To find length of track to be lifted (T, T 2 ) 

Length of circular curve = 2000 x 40°30' rad = 1413-72 ft 

2 + (R + s)tany 

2000 



Tangent length 
new shift 



T / = 4 

3 



S * ' S ' * 1936-5 



- 2-SSx,^ = 2-64 
1936*5 

T 3 1 = 175*0 + (1936-5 + 2*64) tan 20° 15' 

= 175-0 + 715-39 = 890-39 ft 

but 7;/ = 737-84 ft 

.-. TJ 3 = 152-55 ft 

therefore total length of track to be lifted 

= length of arc + 2 x T, T 3 

= 1413-72 + 305-10 = 1718-8 ft 



644 



SURVEYING PROBLEMS AND SOLUTIONS 



To find the length of track to be laid 

= 2 x transition curve (T 3 fj ) + circular arc (f| P z ) 
= 2 x 350 + 2 x 1936-5/3 
= 700 + 3873-0 x 0*26306 
= 700 + 1018-83 = 1718-8 ft. 

12.13 Transition Curves Applied to Compound Curves 

In this case, the transition would be applied at the entry and exit 
(i.e. at T, and T 2 (Fig. 12.31) 

The amount of superelevation cannot be designed to conform to 
both circular arcs and the design speed must relate to the smaller radius. 




Fig. 12.31 Transition curves applied to compound curves 



If the two curves are to be con- 
nected by a transition curve of 
length /, the shortest distance c 
(Fig. 12.32) between the two circu- 
lar curves is given by Glover* as 

c '^(w,~w) (12109 > 

The length of the transition 
L, is bisected at Q by this shift 
c and the shift is bisected by the 
transition. 



^-- * 




Fig. 12.32 



♦Transition curves for railways, Proc. Inst. Civ. Eng., Vol. 140. 



VERTICAL AND TRANSITION CURVES 



645 



If transitions are applied to reverse curves, the radii must be re- 
duced to allow the transition curves to be introduced, Fig. 12.33. 




Fig. 12.33 Transitions applied to reverse curves 

/,/ 2 = /,T 2 + T Z I 2 

= (R, +s,) tan§A, + §L, + \L 2 + (R 2 + s 2 ) tan §A 2 (12.110) 

If L = yJR (based on Gunter chains), 



L 2 
24R 



R 
24/? 



— chains (12.111) 
24 



Shift s, = s 2 

•'• A/ 2 = (*?, + ^) tan |A, + yR, + \y/R 2 + (R 2 + ^) tan \ A 2 

(12.112) 
If R, = R 2 , then 

Uh = (/? + 2^)(tan|A,+tan|A 2 ) + y/R (12.113) 

This may be solved as a quadratic in ^R 

Example 12.7. Two railway lines have straights which are deflected 
through 70°. The circular radius is to be 1500 feet with a maximum 
superelevation of 5 in. The gradient of the line is to be 1 in. in 1 chain 
(Gunter) . 

Calculate the distance from the beginning of the transition to the 
intersection point (i.e. tangent length), the lengths of the separate 
portions of the curve and sufficient data for setting out the curve by 



646 SURVEYING PROBLEMS AND SOLUTIONS 

offsets from the tangent and by the method of tangential angles. 




Fig. 12.34 



Gradient is 1 in. in 66 ft, therefore for the total superelevation it is 5in. 



Shift = 



66 ft = 330 ft 
330 2 



24R 24 x 1500 
3-03 ft 



<*. 



T.I = 1503-03 tan 35° + M 

2 

= 1217-43 ft 

tan" 1 A- = tan" 1 -^ = 6° 16' 37" 
2R 3000 



j8 = 35°- 6° 16' 37" .= 28° 43' 23" 
Length of circular curve = 1500 x 57°26'46;' ad = 1503*93 ft 



Offsets from tangents 



y = 



6RL 



Let the curve be subdivided into 5 equal parts. (1 chain each.) 

66 3 



y^ = 



6 x 1500 x 330 



y 2 = 0-0968 x 2 Z 
y 3 = 0-0968 x 3 J 

y 4 = 0-0968 x 4 ; 



= 0-096 8 ft 

= 0-77 ft 

= 2-61 ft 

= 6-19 ft 



VERTICAL AND TRANSITION CURVES 



647 



y_ = 0-0968 x 5 3 = 12*10 ft 



Tangential angles, based on 1 chain chords. 



a, = 



a 2 
a, 
a A 
a. 



206 265 c 2 206 265 x 66' 



6RL 

= 4a, 
= 9a, 
= 16a, 
= 25a, 

= 3^ 



~ 6x1500x330 

= 1210" 

= 2722-5" 

= 4840" 

= 7562-5 

= 2° 05' 32" 

error 



= 3025" 

(05' 03") 
= 20' 10" 

= 45' 23" 
= 80' 40" 
= 126 '03" 
= 125' 32" 
31" 



Example 12.8. An existing circular curve of 1500 ft radius is to be 
improved by sharpening the ends to 1300 ft radius and inserting a trans- 
ition curve 400 ft long at each straight. 

Using the cubic parabola type, calculate: 

(a) the length of curve to be taken up, 

(b) the movement of the tangent points, 

(c) the offsets for the quarter points of the transition curves. 

(L.U.) 





V 


y\2od 


02 "^ -^^ 








K ^ 










*^\\ 










\ \l300' 






^w 


B 








-_ D 


>* 




~-- 






■ 





Fig. 12.35 



This is Case (4) in Section 12.11. 



In Fig. 12.35, 



PN = y = 



6R 



400' 



1300 



= 20-51 ft 



648 SURVEYING PROBLEMS AND SOLUTIONS 

cf> m = tan-'i^ 
2R 



400 

= Lelll ' 

2 : 
In triangle 2 BP, 



= tan-' Z^y = 8°48' 

2 x 1300 



2 B = 1300cos8°48' = 1284-66 ft 

BP = FN = 1300 sin 8° 48' = 198*90 ft 

0,V = 0,7, - WT, -WV= OJi - PN - 2 B 

= 1500 - 20-51 - 1284-66 = 194-83 ft 

In triangle 0,0 2 V , 

X = cos" 1 194 ' 83 = 13° 05' 
200-0 

V0 2 - 200sinl3°05' =45-28 ft 

(a) Length of curve taken up = 1500 x 13°05; ad = 342-5 ft. 

(b) Movement of the tangent points, i.e. distance T y T 2 : 

FN = T 2 F = 198-90 ft 

T y F = V0 2 = 45-28 ft 

T,T 2 = T 2 F - T X F = 153-62 ft 
i.e. along straight from T, . 



(c) Offsets to transition: 

I 3 

y = 



r 



v i = 



6RL 

100 3 



= 0-320 5 ft = 0-32 ft 



6 x 1300 x 400 

y 2 = 2 3 y, = 8y, = 2-56 ft 

y 3 = 3 3 y, = 27y, = 8-66 ft 

y 4 = 4 3 y, = 64y, = 20*51 ft check 



Example 12.9. AB, BC and CD are three straights. The length of 
BC is 40 Gunter chains. BC deflects 60° right from AB and CD 45° 
left from BC. Find the radius r for two equal circular curves, each 
with transition curves of length y/r at both ends to connect AB and 
CD. BC is to be the common tangent without intermediate straight. 
Find also the total length of curve. (L.U.) 



VERTICAL AND TRANSITION CURVES 649 

By Eq. (12.112), 

40-0 = (r +2^) (tan 30° + tan 22° 30') + V 
= (r+ 2 ^-)(0-5774+ 0-414 2) + yjr 
i.e. 0-991 6r + y/r - 39-96 = 

Solving this quadratic equation in yV, i.e. let / = r, 

1=1 = -1 ± V(l + 4 x 0-9916x39-96) 
V 2 x 0-9916 

/ = 5-864 chains 
r = 34-39 chains 
206 265 L _ 206 265^ _ 103132 x 5-864 
9max 2 R 2r 68-78 

= 8793" 
= 2° 26' 33" 
/8, = 30 - 2° 26' 33" = 27° 33' 27" 

Length of circular arc, A x = 2 x 34-39 x 27° 33' 27 r " ad 

= 68-78 x 0-48096 = 33-08 ch 
ft, = 22° 30' - 2° 26 ' 33" = 19° 33' 27" 

A 2 = 2 x 34-39 x 19° 33' 27" 

= 68-78 x 0-34133 = 23-48 ch 
Total length - A, + A z + 2L 

= 33-08 + 23-48 + 11*73 
= 68-29 chains 

Exercises 12(b) 

14. A road curve of 2000 ft radius is to be connected to two straights 
by means of transition curves of the cubic parabola type at each end. 
The maximum speed on this part of the road is to be 70 mile/h and the 
rate of change of radial acceleration is 1 ft/s 3 . The angle of intersec- 
tion of the two straights is 50° and the chainage of the intersection 
point is 5872-84 ft. 
Calculate: 

(a) the length of each transition curve, 

(b) the shift of the circular arc, 

(c) the chainage at the beginning and the end of the composite 
curve, 

(d) the value of the first two deflection angles for setting out the 



650 SURVEYING PROBLEMS AND SOLUTIONS 

first two pegs of the transition curve from the first tangent point as- 
suming that the pegs are set out at 50 ft intervals. 

(I.C.E. Ans. (a) 542-1 ft (b) 6-10 ft 

(c) 4666-33 ft; 6955-93 ft 

(d) 44", 3' 39") 

15. Two tangents which intersect at an angle of 41° 40' are to be 
connected by a circular curve of 3000 ft radius with a transition curve 
at each end. The chainage of the intersection point is 2784 + 26. The 
transition curves are to be of the cubic parabolic type, designed for a 
maximum speed of 60 mile/h and a rate of change of radial acceleration 
is not to exceed 1 ft/s 3 . 

Find the chainage of the beginning and end of the first transition 
curve and draw up a table of deflection angles for setting out the curve 
in 50 ft chord lengths, chainage running continuously through the tan- 
gent point (I.C.E. Ans. 2771 + 70-5; 2773 + 97-7; 

40"; 5'08"; 13' 54"; 26' 42"; 43' 18" ) 

16. The limiting speed around a circular curve of 2000 ft radius calls 
for a superelevation of 1/24 across the 30 ft carriageway. Adopting the 
Ministry of Transport's recommendation of a rate of 1 in 200 for the 
application of superelevation along the transition curve leading from 
the straight to the circular curve, calculate the tangential angles for 
setting out of the transition curve with pegs at 50 ft intervals from the 
tangent with the straight. 

(I.C.E. Ans. 02' 52"; 11' 28"; 25' 48"; 45' 52"; 1°11'40") 

17. Two straights of a proposed length of railway track are to be 
joined by a circular curve of 2200 ft radius with cubic parabolic transi- 
tions 220 ft long at entry, and exit. The deflection angle between the 
two straights is 22° 38' and the chainage of the intersection point on 
the first straight produced is 2553-0 ft. Determine the chainages at the 
ends of both transitions and the information required in the field for 
setting out the midpoint and end of the first transition curve. 

If the transition curve is designed to give a rate of change of 
radial acceleration of 1 ft/s 3 , what will be the superelevation of the 
outer rail at the midpoint of the transition, if the distance between the 
centres of the rails is 4 ft 11 in. ? 

(I.C.E. Ans. 2002-6 ft; 2222-6 ft; 2871-6 ft; 3091-6 ft; 
Offsets 0-46 ft and 3-67 ft; 2-2 in.) 

18. A transition curve of the cubic parabola type is to be set out from 
a straight centre line. It must pass through a point which is 20 ft away 
from the straight, measured at right-angles from a point on the straight 
produced, 200 ft from the start of the curve. 

Tabulate the data for setting out a 400 ft length of curve at 50 ft 
intervals. 



VERTICAL AND TRANSITION CURVES 



651 



Calculate the rate of change of radial acceleration for a speed of 
30 mile/h. 

(L.U. Ans. 0°20'20", 1°26'00", 3° 13' 20", 5° 42' 40", 8°52'50", 
12° 40' 50", 17° 01' 40", 21° 48' 05"; 1-28 ft/s 3 ) 

19. Two straight portions of a railway line, intersecting at an angle 
of 155°, are to be connected by two cubic parabolic transition curves, 
each 250 ft long and a circular arc of 1000 ft radius. 

Calculate the necessary data for setting out the curve using chords 
50 ft long. 

(R.I.C.S./M Ans. shift 2-61 ft; tangent length 4647-9 ft; 0°05'40", 

0°23'00", 0°51'30", 1°31'40", 2° 23' 10") 

20. Two straights of a railway with 4' SVi gauge, intersect at an 
angle of 135°. They are to be connected by a curve of 12 chains radius 
with cubic parabolic transitions at either end. 

The curve is to be designed for a maximum speed of 35 mile/h with 
a rate of gain of radial acceleration of 1 ft/s 3 . 

Calculate (a) the required length of transition, 

(b) the maximum super elevation of the outer rail, 

(c) the amount of shift required for the transition, and 

(d) the lengths of the tangent points from the intersec- 
tion of the straights. 

(R.I.C.S./M Ans. 170-7 ft, 5-8 in., 1-53 ft, 2001-1 ft) 

21. Two railway straights, having an intersection (deviation) angle 
of 14° 02' 40", are to be connected by a circular curve of radius 2000 ft 
with spiral transitions at each end. 

(a) Calculate the superelevation for equilibrium on the circular 
arc, if the design speed is 45 mile/h, g = 32-2 ft/s 2 and the effective 
gauge between rails = 5 ft and thence, 

(b) if this super elevation is introduced with a gradient of 1 in 600, 
what is the length of each transition and of the circular curve. 

(c) Hence, given the point of intersection of the straights, compute 
all data required for setting out one of the spirals by means of deflec- 
tion angles and 50 ft chords. 

(R.I.C.S./L Ans. 4 in., 200ft, 3' 35", 14' 19", 32' 13", 57' 17") 

22. (a) Calculate the setting out data for a circular curve, radius 500 
ft joining two straights with a deviation angle of 30°00'00". 

(b) Show that a curve having a polar deflection angle equal to 
one third of its tangent deflection angle is a lemniscate. 

For the lemniscate, the ideal transition curve relationship between 
length of curve and radius of curvature does not hold. Show why this is 
not usually important. (N.U.) 



652 SURVEYING PROBLEMS AND SOLUTIONS 

23. The curve connecting two straights is to be wholly transitional 
without intermediate circular arc, and the junction of the two transi- 
tions is to be 16 ft from the intersection point of the straights which 
deflects through an angle of 18°. 

Calculate the tangent distances and the minimum radius of curva- 
ture. If the superelevation is limited to 1 vertical to 16 horizontal, 
determine the correct velocity for the curve and the rate of gain of 
radial acceleration. 

(L.U. Ans. 304-2 ft; 958-3 ft; 30mile/h; 0-292 ft/s 3 ) 

24. The superelevation on a road 50 ft wide is to be 3 ft. Calculate 
the radius for a design speed of 40 mile/h and then give the data for 
setting out the curve if the two straights have a deflection angle of 
30°. Transition curves 300 ft long will be applied at each end, but 
the data for setting out of these is not required. 

(L.U. Ans. Rad 1781*5 ft; tangent length 627'9 ft; 
shift 2-10 ft; Circular arc 632-8 ft) 

25. Two straights of a road 20 ft wide intersect at a through chain- 
age of 8765-9 ft, the deflection angle being 44° 24'. The straights are 
to be connected by a circular arc of radius 900 ft with cubic parabolic 
transitions at entry and exit. The curve is to be designed for a speed 
of 45 mile/h, with a rate of gain of radial acceleration of 2 ft/s 3 . De- 
termine the required length of the transition and the maximum super- 
elevation of the outer kerb. Tabulate all the necessary data for setting 
out the first transition with pegs at every 50 ft of through chainage. 

(L.U. Ans. L = 159-7 ft; c = 3-0 ft; Chainage T, - 8318-3 ft; 

Offsets 0-04, 0-63, 2-65, 4-72 ft) 

26. Assuming an equation A = /(<£) where A and are the intrinsic 
co-ordinates of any point on a transition spiral, prove that A 2 = 2RL<t>, 
where R = minimum radius of curvature; L = length of spiral, <p -.= 
the spiral angle. 

A curve on a trunk road is to be transitional throughout with a total 
deviation of 52° 24'. The design speed is to be 60 mile/h, the maximum 
centripetal ratio 0-25, and the rate of change of radial acceleration 
1 ft/s 3 . 

Calculate (1) the length of each spiral, 

(2) the minimum radius of curvature, 

(3) the tangent distance. 

(L.U. Ans. L = 879-8 ft; R = 962 ft; T,/ = 926-6 ft) 

27. A suburban road 30 ft wide, designed for a maximum speed of 40 
mile/h is to deflect through 38° 14' with a radius of 1600 ft. A cubic 
parabola transition is required, with a rate of gain of radial accelera- 
tion of 1 ft/s 3 . 

Calculate (a) the maximum superelevation of the outer kerb, 



VERTICAL AND TRANSITION CURVES 653 

(b) the length of the transition, 

(c) the chainage of the tangent points if the forward chainage of 
the intersection point is 5829*60 ft, 

(d) the chainages of the junctions of the transition and circular 
arcs. (N.R.C.T. Ans. (a) 2-0ft (b) 126-22 ft 

(c) 5211-77; 6405-69 ft 

(d) 5337-99; 6279-47 ft) 

28. The co-ordinates of three points, K,L and M are as follows: 

Point North (ft) East (ft) 

K 700 867 

L 700 1856 

M 1672 2031 

These points define the direction of two railway straights JK(M) 
and LM, which are to be connected by a reverse curve formed by circu- 
lar arcs of equal radius. 

The circular arcs are to be linked together and to the straights by 
easement curves of length (in Gunter's chains) equal to y/R where R 
is the radius of the circular arcs in chains. Calculate the radius of the 
circular arcs. (L.U. Ans. 9-87 chains) 

29. A road 30 ft wide is to turn through an angle of 26° 24' with a 
centre line radius of 600 ft, the forward chainage of the intersection 
point being 3640-6 ft. A transition curve is to be used at each end of 
the circular curve of such a length that the rate of gain of radial ac- 
celeration is 1 ft/s 3 when the speed is 30 mile/h. Find the length of 
the transition curve, the banking of the road for this speed, the chain- 
age of the beginning of the combined curve, and the angle to turn off 
these for the peg at 3500 ft. 

(L.U. Ans. 142 ft; 2'99ft; 3428-6 ft; 0°34'20") 

30. A road curve of 2000 ft radius is to be connected by two straights 
by means of transition curves of the cubic parabola type at each end. 
The maximum speed on this part of the road is to be 70 mile/h and the 
rate of change of radial acceleration is 1 ft/s 3 . The angle of intersec- 
tion of the two straights is 50° and the chainage at the intersection 
point is 5872-84 ft. 

Calculate: 

(a) the length of each transition curve, 

(b) the shift of the circular arc, 

(c) the chainage at the beginning and end of the composite curve, 

(d) the value of the first two deflection angles for setting out the 
first two pegs of the transition curve from the first tangent point, as- 
suming that the pegs are set out at 50 ft intervals. 

(I.C.E. Ans. (a) 541ft, (b) 6-10 ft (c) 4666-95 ft; 6953-25 ft 

(d) 35"; 3' 40") 



654 SURVEYING PROBLEMS AND SOLUTIONS 

31. A circular curve of radius 700 ft and length 410*70 ft connects 
two straights of railway track. In order that the track may be modern- 
ised to allow for the passage of faster traffic and induce less track 
wear, the whole curve and certain lengths of the connecting straights 
are to be removed and replaced by a new circular curve of radius 2500 
ft, with transitions of the cubic parabola type at entry and exit. 

Given that the maximum speed of the traffic on the new curve is to 
be 60 mile/h, and the rate of change of radial acceleration is not to 
exceed 0'90 ft/s 3 , determine: 

(a) the length of the new composite curve, 

(b) the length of the straight track to be removed 

(c) the necessary superelevation of the track on the circular curve, 
the gauge of the track being 4 ft 8^ in. 

(I.C.E. Ans. (a) 1769-7 ft; (b) 2 -695-7 ft; (c) 5-42in) 

Bibliography 

CLARK, D., Plane and Geodetic Surveying, Vol. 1 (Constable) 
BANISTER, A. and Raymond, s., Surveying (Pitman) 
JAMESON, A.H., Advanced Surveying (Pitman) 
JENKINS, R.B.M., Curve Surveying (Macmillan) 

searles, w.H. and IVES, H.C., ed. by p. KIRSAM, Field Engineering, 

22nd Ed. (John Wiley) 
HIGGINS, A.L., Higher Surveying (Macmillan) 
SHORTT, W.H., A Practical Method for Improvement of Existing 

Railway Curves, Proc. Inst. Civ. Eng., Vol. 176 
ROYAL-DAWSON, F.G., Elements of Curve Design for Road, Railway 

and Racing Track (Spon,1932) 
ROYAL-DAWSON, F.G., Road Curves for Safe Modern Traffic (Spon, 1936) 
PERROTT, S.W. and BADGER, F.E.G., The Practice of Railway Survey- 
ing and Permanent Way Work (Edward Arnold, 1920) 
SHARMA, R.C. and SHARMA, S.K., Principles and Practice of Highway 

Engineering (Asia Publishing House) 



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Solutions 



$8 









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